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I managed to get the same result, but I didn't want to post because I couldn't figure out why I was calculating angle due East instead of North. The solution clarified it, definitely need to brush up on my trig! | 677.169 | 1 |
$\begingroup$@YvesDaoust How is it incorrect? Given that I am given a constrained triangulation, the edges in the triangulation are either fully inside or fully outside the polygon. Testing the midpoint is 100% correct. If any point is inside then the entire edge is inside.$\endgroup$
$\begingroup$It is not clear from the question that the segments are either wholly in or wholly out. If this is the case, why taking the midpoint rather than one of the endpoints ? And what is the problem then, point-in-polygon is old as the world.$\endgroup$
– user16034
Aug 13, 2023 at 14:21
$\begingroup$@YvesDaoust I explicitly said I am working on the segments of a constrained triangulation, if you look at any constrained triangulation that has a closed polygon as the constraint no edge will cross the polygon, by construction of the triangulation. You use the midpoint because the endpoints may lie right at the boundary, this would lead to potential numerical imprecision, the midpoint is guaranteed to lie in the interior so it will perform much better.$\endgroup$
Finding the First Triangle
To find the first triangle inside of the polygonal constraint, You could simply take one edge of the polygon $(a,b)$ and find the triangle $(a,b,c)$, provided the polygonal constraint has the same orientation as the triangles.
Finding The Triangle
In order to find the triangle that contains $a$ You could build some BVH with all the triangles and in $\mathcal{O}(n*\log(n))$ and find a triangle in $\mathcal{O}(\log(n))$. But that may not be as fast in practice as in theory.
Another way is to start at some triangle and then walk along the mesh towards $a$ using the A* algorithm. If You sort the tested edges according to some space-filling curve order, and You always start with the triangle You've last looked at, You may even find the next triangle in amortized $O(1)$ time in practice. | 677.169 | 1 |
my-frugal-lifestyle
Draw two isosceles triangles, ∆ABC and ∆ADC with common base AC . Vertexes B and D are in the oppos...
4 months ago
Q:
Draw two isosceles triangles, ∆ABC and ∆ADC with common base AC . Vertexes B and D are in the opposite semi-planes determined by AC . Draw the line segment BD . Prove: AC ⊥ BD
Accepted Solution
A:
Let ΔABC is an isosceles triangle. AB = BC and AC is non-equal side of the triangle. Let draw the median of side AC , which meets at vertex B and bisect ∠B and bisect AC . We can mark mid point of AC as 'O'. So AC⊥BO. Similarly ΔADC is an isosceles triangle. Median of DC meets and bisect ∠D.The mid point of ADC of side AC is O. Then AC⊥ DO. When we join B and D It passes through mid point of AC. Therefore AC⊥BD. | 677.169 | 1 |
45 Geometry Chapter 4 Congruent Triangles Answer Key
Geometry Chapter 4 Congruent Triangles Answer Key
Introduction
Geometry can be a challenging subject for many students, especially when it comes to understanding the concept of congruent triangles. In Chapter 4 of the Geometry textbook, students are introduced to the properties and theorems related to congruent triangles. This article aims to provide an answer key for Chapter 4 of the Geometry textbook, specifically focusing on congruent triangles.
1. Definition of Congruent Triangles
Before diving into the answer key, it is essential to have a clear understanding of what congruent triangles are. Congruent triangles are two triangles that have the same shape and size. In other words, all corresponding sides and angles of congruent triangles are equal.
2. Congruence Criteria
There are several criteria for proving that two triangles are congruent. These criteria include Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side (AAS), and Hypotenuse-Leg (HL). Understanding these criteria is crucial for solving congruent triangle problems.
3. SSS Congruence Postulate
The SSS Congruence Postulate states that if the three sides of one triangle are congruent to the corresponding sides of another triangle, then the two triangles are congruent. This postulate is often used to prove congruence using measurements.
4. SAS Congruence Theorem
The SAS Congruence Theorem states that if two sides and the included angle of one triangle are congruent to the corresponding parts of another triangle, then the two triangles are congruent. This theorem is commonly used to prove congruence when angle measurements are involved.
5. ASA Congruence Theorem
The ASA Congruence Theorem states that if two angles and the included side of one triangle are congruent to the corresponding parts of another triangle, then the two triangles are congruent. This theorem is often used to prove congruence when side lengths are involved.
6. AAS Congruence Theorem
The AAS Congruence Theorem states that if two angles and a non-included side of one triangle are congruent to the corresponding parts of another triangle, then the two triangles are congruent. This theorem is frequently used to prove congruence when angle measurements are provided.
7. HL Congruence Theorem
The HL Congruence Theorem, also known as the Hypotenuse-Leg Congruence Theorem, states that if the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of another right triangle, then the two triangles are congruent. This theorem is specifically used when dealing with right triangles.
8. Example Problems
In this section, we will provide step-by-step solutions to a few example problems involving congruent triangles. These solutions will help students understand how to apply the congruence criteria and theorems discussed earlier.
13. Problem 5 - Applying HL Congruence Theorem
Solution: By using the HL Congruence Theorem, we can conclude that triangle XYZ is congruent to triangle UVW.
14. Additional Concepts
Aside from the congruence criteria and theorems, there are other concepts related to congruent triangles that students should be familiar with. These concepts include corresponding parts of congruent triangles, congruence shortcuts, and the concept of congruence in real-world scenarios.
15. Corresponding Parts of Congruent Triangles
When two triangles are congruent, their corresponding parts are also congruent. This means that the corresponding angles and sides of congruent triangles are equal in measure.
16. Congruence Shortcuts
There are several shortcuts or properties that can be used to determine congruence without explicitly proving each criterion. These shortcuts include the Reflexive Property, Symmetric Property, and Transitive Property.
17. Congruence in Real-World Scenarios
Congruence is not just limited to abstract geometric shapes. It can be observed and applied in real-world scenarios as well. For example, architects and engineers use the concept of congruence to ensure that structures are stable and balanced.
18. Conclusion
Chapter 4 of the Geometry textbook covers the topic of congruent triangles, which is essential for understanding geometric concepts and problem-solving. This article provided an answer key for Chapter 4, highlighting the congruence criteria and theorems, as well as example problems to illustrate their application. By mastering the concepts of congruent triangles, students can enhance their geometric reasoning and problem-solving skills.
19. Additional Resources
For further practice and reinforcement of the concepts covered in Chapter 4, students can refer to additional resources such as online tutorials, practice worksheets, and interactive geometry software.
20. References
List of references used in the creation of this answer key for Chapter 4 of the Geometry textbook. | 677.169 | 1 |
The Elements of Euclid; viz. the first six books, together with the eleventh ...
BOOK VI. is parallel to KC, one of the sides of the triangle DKC, as CE to ED, 2. 6. so is KH to HD: But KH is equal to BG, and HD to GF; therefore, as CE to ED, so is BG to GF: Again, because FD is pa- G rallel to EG, one of the sides of the triangle AGE, as ED to DA, B so is GF to FA: But it has been
D
HE
K
proved that CE is to ED, as BG to GF; and as ED to DA, so GF to FA: Therefore the given straight line AB is divided similarly to AC. Which was to be done.
PROP. XI. PROB.
To find a third, proportional to two given straight lines.
Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; A it is required to find a third proportional
to AB, AC.
Produce AB, AC, to the points D, E; and make BD equal,to AC; and having joined BC, through D, draw DE parallel a 31. 1. to it a.
B
Because BC is parallel to DE, a side of 12. 6. the triangle ADE, AB is to BD, as AC D
E
to CE: But BD is equal to AC; as therefore AB to AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Which was to be done.
PROP. XII. PROB.
To find a fourth proportional to three given straight lines.
Let A, B, C be the three given straight lines; it is required to find a fourth proportional to A, B, C.
Take two straight lines DE, DF, containing any angle Book VI.
but DG is equal to A, GE to B, and DH to C; therefore, as A is to B, so is C to HF. Wherefore to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done.
b 2.6.
PROP. XIII. PROB.
To find a mean proportional between two given straight lines.
Let AB, BC be the two given straight lines: it is required to find a mean proportional between them.
Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and
from the point B draw a BD at right angles to AC, and join AD, DC.
Because the angle ADC in a semicircle is a right angleb, and because in the right
angled triangle ADC, BD is A
D
B
11. 1.
b 31. 3.
drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base: Therefore between the two given straight lines Cor. 8. 6. AB, BC, a mean proportional DB is found. Which was
to be done.
с
M
BOOK VI.
a
PROP. XIV rallelogram AB is equal to A
pa
b
F
BC, and that FE is another
parallelogram, AB is to FE,
7.5.
as BC to FEb: But as AB
to FE, so is the base DB to
B
G
C
1. 6. BE: and as BC to FE, so is the base of GB to BF; therefore, as DB to BE, so is
11. 5. GB to BFd CB to BF, so is the parellelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FEd: 9. 5. Wherefore the parallelogram AB is equale to the parallelogram BC. Therefore equal parallelograms, &c. Q. E.D.
BOOK VI.Let the triangles be placed so, that their sides, CA, AD be in one straight line; wherefore also EA and AB are in one straight linea; and join BD. Because the triangle * 14. 1. ABC is equal to the triangle
:
ADE, and that ABD is another triangle
D
so is the base EA to AB: as therefore CA to AD, so is EA to AB BAC to triangle BAD, so is triangle EAD to triangle BAD; that is, the triangles BAC, EAD have the same ratio to the triangle BAD: Wherefore the triangle ABC is equal to the tri- 9. 5. angle ADE. Therefore equal triangles, &c. Q. E. D.
BOOK VI.
11. 1.
PROP. XVI. THEOR.
Ir drawa. is equal to CH, and 7. 5. F to AG; AB is to CD as CH to AG. Therefore the sides of the parallelograms BG, DH about the equal angles. are reciprocally proportional; but parallelograms which have their sides about equal angles reciprocally propor14. 6. tional, are equal to one another; therefore the parallelogram BG is equal to the parallelogram DH: And the parallelogram BG is con- E
tained by the straight lines AB, F; because AG is equal to F; and the pa- rallelogram DH is con- tained by CD and E; be- cause CH is equal to E: Therefore the rectangle contained by the straight lines AB, F is equal to
F
G
A
that which is contained by CD and E.
"And if the rectangle contained by the straight lines, AB, F be equal to that which, the rectangle BG is contained by AB, F, because AG is equal to F; and the rectangle DH by CD, E, because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram | 677.169 | 1 |
Choose the correct answer Let a and b be two unit vectors and θ is the angle between them. Then, a + b is a unit vector, if
a) θ=π4 b) θ=π3 c) θ=π2 d) θ=2π3
Open in App
Solution
Let a and b be two unit vectors and θ be the angle between them. Then, |a|=|b|=1 Now, (a + b) is a unit vector if |a+b|=1⇒(a+b)2=1⇒(a+b).(a+b)=1⇒a.a+a.b+b.a+b.b=1 ⇒|a|2+|b|2+2a.b=1[∵a.b=b.a] ⇒12+12+2a.b=1⇒a.b=−12⇒|a||b|cosθ=−12 ⇒1×1×cosθ=−12⇒cosθ=−12⇒θ=2π3 | 677.169 | 1 |
...straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained...and the several parts of the divided line, Let A and .BCbe two straight lines ; and let BC be divided into any parts SD, DE, EC, in the points D, E. Then...
...straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained...undivided line, and the several parts of the divided line. Given the two straight lines A and BC, and let BC be divided into any parts in the points D, E ; to...
...straight lines, one of which is divided into any number of parts ; then the rectangle contained by the two straight lines, is equal to the rectangles contained...undivided line, and the several parts of the divided line. (References— Prop. i. 3, 11, 31, 34.) Let A and BC be two straight lines; and let BC be divided into...
...of which is dicided into any number of parts, the rectangle contained by the tiro straight lin<!,< is equal to the rectangles contained by the undivided line, and the several parts of the divided line. , (References— Prop. I. 3, 11, 31, 34.) Hypothesis. — Let A and BC be two straight lines ; and...
...straight lines, one of which is divided into any number of parts ; the rectangle contained by the two straight lines is equal to the rectangles contained...undivided line and the several parts of the divided line. 8 If a straight line be divided into two equal parts, and also into two unequal parts ; the rectangle...
...straight lines, one of irhich is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained...straight lines ; and let BC be divided into any parts BD, DE, EC, in ftfepoints D, E. Then the rectangle contained by the straight lines -flfcd BC, shall...
...straight lines, one of tchich is divided into any number of parts ; the rectangle contained by the two straight lines, is equal to the rectangles contained...and the several parts of the divided line. Let A and BCbe two straight lines ; and let BC be divided into any parts BD, DE, EC, in the points D, E. Then...
...straight lines, one of which is divided into any number of parts ; the rectangle contained by the two straight lines is equal to the rectangles contained...undivided line, and the several parts of the divided line. Prop. 2. If a straight line be divided into any two parts, the rectangles contained by the whole and...
...straight lines, one of which is divided into any number of parts ; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several farts of the divided line. Let A and JBC be two straight lines ; and let BC be divided into any parts...
...into any number of parts, the rectangle contained by the two lines is equal to the sum of the several rectangles contained by the undivided line and the several parts of the divided line. Let AB and AD be two lines, and suppose AB divided into any number of parts at the points E, F, G, etc.... | 677.169 | 1 |
Angles
What are angles? An angle starts with two straight lines that meet at a common point. The point is called a vertex. The Latin word 'angulus,' means "corner." The measure of degrees between the two lines is considered the angle. For example, if you stand up in a room in your house, the intersection of the plane of the floor and the line of you standing on it create a 90-degree angle. If you stand with your legs and arms apart in the shape of an "x", your legs create an angle and the space between your arms does, too. Look down at your hand. Open your fingers. Between each two fingers is an angle! Find some scissors. Open the scissors a little and you will see an angle between the two blades; open the scissors wider and you will a wider angle! Find a wall clock or a watch. You will see the hour hand and the minute hand as they tick away the time. They form an angle. Angles are all around us! Become an angle explorer.
Activity 1 – Measuring Angles
As you discovered in the introduction, angles can be narrow and they can be wide. The flat floor is actually a 180-degree angle. If you lie down on the floor and lift your legs straight up, you will form a 90-degree angle. If you lay down, someone can draw a half circle starting by at the end point of your outstretched arms over your waist and to your feet and create a 180-degree circle. If you flap your arms and legs open and closed (like making a snow angel) you can mark a complete circle of 360 degrees.
Think again of the clock. The hands move 360 degrees around a circle. If you divide the circle into half you will have two 180-degree half circles; if you divide the half circles into half you will have four quarters or four 90-degree angles. Now divide the circle into thirds, fifths, and sixths. Using a protractor can help you divide a circle into halves, thirds, fourths, and so on. The equation to find the angle degrees for each of these will be the angle in degrees = 360 degrees of a circle divided by 3 (120 degrees), divided by 5 (72 degrees), and divided by 6 (60 degrees). Draw 4 circles divided in half, divided in 4ths, divided in 5ths, and divided in 6ths. Upload them to the gallery.
Activity 2 – Types of Angles
If an angle is narrower than 90 degrees, it is called an acute angle. If it is between 90 and 180 degrees it is called an obtuse angle. If it is 180 degrees, it is called a straight angle. If it is between 180 and 360 degrees it is called a reflex angle. If it is 360 degrees it is called a complete angle. Draw and label all of the types of angles. A reflex angle is an angle that measures between 180˚ and 360. Draw and label an acute angle, an obtuse angle, a straight angle, a reflex angle, and a full angle. Know your angles!
Activity 3 – Triangles and Inside and outside angles
A triangle is a three-sided polygon. All triangles have three sides and three angles. The triangle's three interior angles all add up to 180 degrees. Think of the pyramids. Draw three different heights of pyramids. Draw one that has a narrow base and two tall sides. Now draw a triangle that has a very wide base and is shorter than the narrow base triangle. Draw another that has three equal sides and three equal angles. This triangle is an equilateral triangle meaning that all sides and angles are equal to each other. Using a protractor, measure the inside and outside angles at each vertex and label them. The one with equal inside angles is called an Equilateral Triangle. A Scalene triangle is a triangle where all sides and all angles are different lengths or different degrees. An Isosceles Triangle is a triangle that has two equal sides of three, and two equal angles of three. Draw three different triangles and label their type.
Review
The point where two straight lines meet together to become an angle is called a vertex.
true
false
360 degrees make up a circle.
true
false
What kind of angle is narrower than 90°? - acute, obtuse, straight, reflex
obtuse
straight
acute
reflex
What kind of triangle is made up of three equal sides? - equilateral, scalene, isosceles | 677.169 | 1 |
Similar Polygons Worksheet Answers
Similar Polygons Worksheet Answers – Visit the reading comprehension web page for a complete assortment of fiction passages and nonfiction articles for grades one by way of six. Enter the price paid by each parent for work-related youngster care. If the price varies , take the whole yearly price and divide by 12. The custodial father or mother is the mother or father who has the kid more of the time. If each of you may have the kid 50% of the time, choose certainly one of you to be the custodial father or mother. Select Text AreaTo select a text area, hold down the or key.
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Create Perfect Similar Polygons Worksheet Answers
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The second type of math worksheet is meant to introduce new matters, and are often completed within the classroom. They are made up of a progressive set of questions that leads to an understanding of the topic to be learned. It is normally a printed page that a toddler completes with a writing instrument.
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If income varies so much from month to month, use a mean of the last twelve months, if obtainable, or final year's income tax return. When you load a workbook from a spreadsheet file, it will be loaded with all its current worksheets . Move on to activities during which college students use the primary sources as historic evidence, like on DocsTeach.org.If you are looking for Similar Polygons Worksheet Answers, you've arrive to the right place. We have some images very nearly Similar Polygons Worksheet Answers including images, pictures, photos, wallpapers, and more. In these page, we after that have variety of images available. Such as png, jpg, busy gifs, pic art, logo, black and white, transparent, etc.
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Honors Geometry – Vintage High School: Section 7-2 Similar Polygons
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You can resume or droop the chosen warehouse, or resize the warehouse. The record of databases and other objects refreshes mechanically when the worksheet context is modified. Users can also click the refresh button on the top of the object browser to view object changes instantly. Click the CMD or CTRL key once and then...
Right Triangle Trig Worksheet. Pleasant for you to my website, in this period I am going to explain to you about Right Triangle Trig Worksheet. What about graphic preceding? can be of which awesome???. if you feel thus, I'l d teach you a number of image again beneath: So, if you'd like to acquire these...
Trim levels has been simplified from back it originally came to bazaar in 2014 to aloof Zetec and Titanium. This agency that there is far added accepted kit on the entry-level versions than before. Opt for the entry-level Zetec archetypal and you will acquisition the Grand Tourneo Connect comes with a manually adjustable driver's seat,... | 677.169 | 1 |
What is the exact value of sin 3pi 4?
One can also ask, what is the exact value of sin 3pi 2? The exact value of sin(π2) sin ( π 2 ) is 1 . Multiply −1 by 1 .
Consequently, what is the exact value of CSC 3pi 4?
The exact value of csc(π4) csc ( π 4 ) is √2 .
What is the exact value of cot 3pi 4?
The exact value of cot(π4) cot ( π 4 ) is 1 . Multiply −1 by 1 .
Associated Query Solutions
What is Cospi?
cos(π)=cos(−π) are the each cos values for the similar place. They're each equal to (−1) as a result of. if considered as a unit circle centered on the Cartesian origin. the cos is the x value.
What is the exact value of sin 5pi 6?
The exact value of sin(π6) sin ( π 6 ) is 12 .
What is the sine value of 7pi 6?
The exact value of sin(π6) sin ( π 6 ) is 12 .
How do you discover sin 2pi 3?
The exact value of sin(π3) sin ( π 3 ) is √32 .
What is the exact value of sin 5pi 3?
The exact value of sin(π3) sin ( π 3 ) is √32 .
How do you discover the exact value of cos 3pi 4?
The exact value of cos 3pi/4 is -1/√2. In a proper triangle, the cosine angle is divide the size of hypotenuse (H) by the size of the adjoining facet (A). The phrase sine comes from the Latin phrase sinus and the prefix 'co' with sine make the cosine phrase.
What is the reference angle of 3pi 2?
How do you exchange from levels to radians?
To convert levels to radians, take the quantity of levels to be transformed and multiply it by π/180. You'll be able to calculate this by changing each numbers into fractions. For instance, to convert 120 levels you'll have 120 x π/180 = 120π/180.
Related Questions:How do you find COT 45 degrees?The worth of cot perform when angle of proper triangle equals to 45∘ known as cot of angle 45 levels. cot(45∘)=1 ? cot(45∘) ? will be written in various kind...
What is the value of Cosec Pi by 4?The precise value of csc(π4) csc ( π 4 ) is √2 . Furthermore, what is the value of Cot Pi by 4? Make the expression unfavorable as a result...What is a social sin examples?Examples of Social sin could embrace Warfare and Poverty. These results injury whole communities and international locations. Moreover, what is meant by social sin? Definition. Social Sin is a private...
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Why is it Mgsin Theta?Forces and Torques The free-body diagram of the item reveals two forces parallel to the slope. If the angle of the incline is theta, the element of the power of | 677.169 | 1 |
Six trigonometric ratios sheet 1 math worksheets 4 kids. Free printable trigonometry worksheets right triangle trigonometry worksheet and inverse trig functions worksheet are some main things we want to show you based on the gallery title. 3 21 cm c.
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Find the measure of each angle indicated. Right triangle trigonometry word problems. Modeling with right triangles. | 677.169 | 1 |
@Ishwaran I had mentioned earlier that the line TS can't be parallel to PQ unless the base angles of original triangle are equal. It is best to draw a random triangle instead of a symmetric figure and then you can see why TS can't be parallel to PQ.
Also you should note that you are supposed to prove similarity of RPQ and RST. While proving similarity and congruence the order of letters is very important and that tell which vertex of one triangle corresponds to which vertex of another triangle. Here P corresponds to S and Q to T
So the angle at P matches angle at S. For parallel lines you should have angle at P equal to angle at T
he did enjoy it. they don't talk very often, so each time there is a lot of vocabulary development that has taken place since last time. as a narrative it was pretty hard to follow. she doesn't fully appreciate how people on the phone aren't in the room, so she changes subjects or emphasizes things by pointing at stuff.
@robjohn I think that would work if $du + i (pdx + dy) = gdz$ held globally in $\Omega$, but this only holds locally in $\Omega$, for not necessarily the same holomorphic $g$ at different points, unless I misunderstand your answer
Question:
" The average marks of the students in four sections A,B,C,D together is 60 %. The average marks of the students of A,B,C,D individually are 45 %,50 %,72% and 80 % respectively. If the average marks of the students of the sections A and B together is 48 % and that of the students of B n...
Oh you cant do that, if you did you would be saying LHS is already equal to RHS.... we are supposed to prove they LHS=RHS but we cant make the assumption they are already equal (at least in these sort of questions)
Yahhi did you take the denominator to the other side? Or anything to the other side, for that matter? the correct method is turning LHS into RHS and then telling "Tada, look I turned LHS into something that looks like the RHS, therefore the equation is true"
@Koro i thought about using that formula, but in his grade its not yet introduced so i decided against using it, but I think you can skip it if I am right (i just did an eyeballing of the solution, i am not sure)
The claim is easy if $A_n \searrow A$, or $A_n \nearrow A$. But if, say, $A_{2n} = [-2, 1]$, and $A_{2n+1} = [-1, 2]$, then I don't even see a set $A$ that satisfies the requirement $\mathbb{P}(A_n \triangle A) < \epsilon$ for any $n$.
sorry, $f: \Sigma \rightarrow \mathbb{R}$.
in fact, the claim follows if $(A_n)_n$ converges, not even necessarily monotonically.
@JoeShmo $\int_{\Sigma} |\mathbf{1}_{A_n} - f| d \mu \rightarrow 0$ implies $\mathbf{1}_{A_n}$ converges to $f$ in measure, since you are working in a probability space (a finite measure space), this implies that there is a subsequence of the $\mathbf{1}_{A_{n_k}}$ converging to $f$ almost everywhere, clearly then, $f$ is equal to $\mathbf{1}_{\liminf_{k} A_{n_k}}$ almost everywhere
pick a ball around $p$, every point in the ball except $p$ has a local orientation. argue these all come from the same orientation on the ball. then, if you equip $p$ with the local orientation induced by the orientation of that ball, they all glue together to an orientation of $M$.
the dimension assumption comes into play when you use the deleted ball is connected in dimensions >1. the claim is still true in dim 1, but for different reasons
I've got a question: Given a big (uncountable) algebraically independent set $A\subseteq\mathbb{R}$, can you say anything about the outer measure of $A$? My instinct tells me that the two are independent topics, and the measure could be trivial or infinite or anywhere in between, but I don't legitimately know
@Thorgott You mean if $p\in B$ is an open ball then there is a unique generator $\mu_B\in H_n(M,M-B)$ such that $i^x_*(\mu_B) = \mu_x$ for each $x\in B\setminus\{p\}$? Here, $\mu_x$ is a local orientation and $i^x:(M,M-B)\to (M,M-x)$ is an inclusion map.
the point is that we know a priori that $H_n(M,M-B)$ is cyclic and $i^x$ induces an iso on $H_n$ for each $x\in B$, so the only potential issue is consistency, but local consistency holds by definition and passing to global consistency is a connectedness argument
Now for the other direction: Since W is FDVS, let $w_1,...,w_k$ be a basis for W. I define $S:W\to W$ by $Sw_i=w_i$ (identity) and then I claim that $T_2=ST_1$. For any $x\in V$ we have $ST_1(x)= T_1(x)$, I am stuck here.
i dunno how to think about this without giving it away. S is gonna have to take stuff that T_1 acted on and move it around so it lines up with what T_2 does. maybe a starting point is to pick a basis for the range of T_1 (if it is nonzero; if it's zero that is an easy special case). and fiddle around with first defining S on that. bear in mind you will need to use the hypothesis about nullspaces somewhere.
@TedShifrin It seems complicated. My experience in linear algebra so far has been in matrices/determinants mostly. It's probably the first time I'm reading linear maps (not necessarily from R^m to R^n, wherein I could write linear map in form of matrix times a column vector).
I read linear algebra for the first time few years back when I was at college. Now, I'm not at college. I started reading linear algebra and after going through vector spaces, now I'm at linear maps (chapter 3 in linear algebra done right).
i do think there can be a problem in the bounded case. T_1 can be injective but also crunch norms of nonzero inputs arbitrarily close to zero (e.g. "multiplication by x" on C[0,1] with inputs supported in tiny neighborhoods of zero). taking T_2 = I, it seems that your S would potentially have to blow the norms of some vectors up by arbitrarily large factors.
on review of that comment, i think he was just being sloppy with notation. but i'm not sure that his operator is closed, because he didn't impose anything resembling boundary conditions on the domain and you usually need to do that. i couldn't figure out if he was in R^1, where it might not make a difference, or R^n.
generally if you say "let the domain be the set of things for which both sides of this formula make sense," where the formula involves a differential operator, you don't get even a closable operator without boundary conditions or regularity conditions. even for a one dimensional candidate for d/dxv=\sum d_i vin many books script letters are often used to denote things that are larger than sets and cannot be sets. that might be what is going on here. what is it a collection of?
if it's in his set theory book, that's probably what he's doing. otherwise, who knows. every author seems to like at least one form of alternate typesetting, whether script or fraktur or who knows what.
Yeah it's from his set theory book. I think he uses lower-case letters to denote sets, and then upper-case letters to denote sets which contains sets, and the weird letters to denote sets which contain sets which contain sets
@leslietownes yes, taking inspiration from what I want to prove ($S_2=ST_1$). Defining S:W$\to W$ as $S(T_1v_i)=T_2v_i$ (for all $1\le i \le k$) determines S completely and uniquely (from a theorem that says that a linear map F:X->Y, where X is FDVS is completely determined by its values on basis of X.).
Now for any v in V, there exist $c_i$'s (scalars) and v' in null T_1 such that $v=c_1v_1+…+c_kv_k+v'$, which gives $T_1v= c_1 T_1v_1+…+ c_k T_1 v_k$ so $ST_1v= T_2(c_1v_1+…+c_kv_k)$. Since v' is in null T1 which is a subset of null T2, it follows that v' is in null T2. So we get $ST_1v = T_2 (c_1v_1+…+c_kv_k)+T_2 (v')= T_2( c_1v_1+…+c_kv_k+ v')= T_2 v$leslietownes ah yes, you're absolutely right. The working above can be fixed if I just remove the word S is uniquely determined. We know S exists (by the existence theorem for linear maps) and that's it.
Also the representation of v in V as the sum $v=(c_1v_1+…+c_kv_k)+v'$ , where v' is in null T1, is unique as sub space spanned by v_i's and the subspace null T_1 have only 0 in common.
i think so. if i were writing this up for future personal reference i might re-order the argument and make the various steps more distinguished from one another. things tend to spill out in fits and starts in a chat window.
news flash, my daughter has announced that after she turns into a cat she is going to turn into an elephant.
@Koro Here is another, more direct way (I think): One direction is immediate. For the other, let $v_1,..,v_p$ be a basis for $\operatorname{im} T_1$ and add $v_k$ as necessary to make a basis for $W$. Suppose $x_1,...x_p$ are such that $v_k = T_1 x_k$ (note that the $x_k$ are linearly independent), and define the linear $S v_k = T_2 x_k$ (and zero on the rest).
copper, interesting piece in the chronicle this morning. an editor was cleaning out some old papers and found letters from the unabomber asking about places to go in south america (the editor had written a book on traveling there). before he began unabombing. | 677.169 | 1 |
Class 8 Courses
Let S = the set of points inside the square S = the set of points inside the square, T = the set of points inside the triangle and C = the set of points inside the circle. If the triangle and circle intersect each other and are contained in a square. Then,
(a) S ∩ T ∩ C = ϕ
(b) S ∪ T ∪ C = C
(c) S ∪ T ∪ C = S
(d) S ∪ T = S ∩ C
Solution:
Let S = the set of points inside the square
T = the set of points inside the triangle
C = the set of points inside circle
Given triangle and circle intersect each other and are contained in a square | 677.169 | 1 |
(a) Look at the following matchstick pattern of squares (see figure). The squares are not separate. Two neighbouring squares have a common matchstick. Observe the pattern and find the rule that gives the number of matchsticks in terms of the number of squares. (b) Figure gives a matchstick pattern of triangles. As in Exercise 11(a) above, find the general rule that gives the number of matchsticks in terms of the number of triangles. | 677.169 | 1 |
Lines and Angles: Geometry Introduction
Today I am start Geometry series. In this series I will share all the basic rules and techniques to solve questions from this chapter.
Complementary Angles:
Two angles whose sum is 90 degree are called Complementary Angles.
For example 40 degree and 50 degree
Their sum =40 degree +50 degree= 90 degree
Supplementary Angles:
Two angles whose sum is 180 degree are called Supplementary Angles.
For example 100 degree and 80 degree
Their sum =100 degree + 80 degree= 180 degree
Linear pairs:
Two angles on a given line are called linear if their sum is 180 degree. Look at the picture below.
In the above figure angle ABD and angle CBD are linear pair because their sum is 180 degree.
Vertically opposite Angles:
Vertically opposite angles are formed when two lines, say AB and CD intersect each other at a point O. There are two pairs of vertically opposite angles and they are always equal.
One pair is angle AOD and angle COB
Angle AOD = Angle COB (Vertically opposite angles)
The other pair is angle AOC and angle BOD
Angle AOC = Angle BOD ( Vertically opposite angles)
Note : If two lines intersect each vertically opposite angles are equal.
Transversal
A line which intersects two or more lines at different points is called a transversal.
Let us understand this:
In the above picture line l intersects lines m and n at points P and Q respectively. Thus line l is a transversal for lines m and n. By observing the picture, we can see that there are four angles formed at each point P and Q.
Alternate Exterior Angles
Interior Angles on the same side of the transvers | 677.169 | 1 |
STD 2nd Sub- Math Mark-20) Answer in one words.5
How many sides does a rectangle have?
A.
One
B.
Two
C.
Three
D.
Four
Correct Answer D. Four
Explanation A rectangle has four sides because it is a quadrilateral with four straight sides and four right angles. Each side of a rectangle is parallel to the opposite side, and the opposite sides are equal in length. Therefore, the correct answer is four.
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2.
How many sides does a triangle have?
A.
Four
B.
Three
C.
Two
D.
One
Correct Answer B. Three
Explanation A triangle has three sides because it is a polygon with three straight sides. The definition of a triangle is a closed figure with three sides and three angles. Therefore, the correct answer is three.
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1
0
3.
How many sides does a square have?
A.
Three
B.
one
C.
Four
D.
One
Correct Answer C. Four
Explanation A square has four sides because it is a quadrilateral with four equal sides. Each side of a square is perpendicular to the adjacent sides, and all angles in a square are right angles. Therefore, the correct answer is four.
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4.
Does a square have any Curved side?
A.
Four
B.
One
C.
Three
D.
No
Correct Answer D. No
Explanation A square does not have any curved side because all sides of a square are straight and equal in length. A curved side would imply a change in direction, which is not present in a square. Therefore, the correct answer is "no".
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5.
Does a circle have any straight side?
A.
no
B.
two
C.
Four
D.
One
Correct Answer A. no
Explanation A circle does not have any straight sides because by definition, a circle is a closed curve with all points equidistant from the center. It is made up of an infinite number of points and has a curved shape, with no straight sides.
Rate this question:
6.
Q-2-A)Fill in the blanks.4
c__ ne
A.
A
B.
B
C.
O
D.
I
Correct Answer C. O
Explanation The missing letter in the word "c__ne" is "o".
Rate this question:
7.
Cyli__Der
A.
D
B.
B
C.
A
D.
N
Correct Answer D. N
Explanation The given string "cyli__Der" has multiple underscores in it. The correct answer "n" is the only character in the string that does not have an underscore before it.
Rate this question:
8.
Cu__oid
A.
I
B.
B
C.
D
D.
A
Correct Answer B. B
9.
cir__le
A.
A
B.
D
C.
C
D.
B
Correct Answer C. C
Explanation The given answer "c" is the correct answer because it completes the word "circle". The missing letter is "c" and when it is added to the word "cir__le", it forms the word "circle".
Rate this question:
10.
B) complete the table.
4×5= .....
A.
23
B.
20
C.
21
D.
39
Correct Answer B. 20
Explanation The correct answer is 20 because when you multiply 4 by 5, you get 20.
Rate this question:
11.
6×9= .....
A.
53
B.
52
C.
54
D.
56
Correct Answer C. 54
12.
7×7=.....
A.
49
B.
39
C.
47
D.
50
Correct Answer A. 49
Explanation The correct answer is 49 because when you multiply 7 by itself, you get 49.
Rate this question:
13.
3×9=.....
A.
28
B.
29
C.
32
D.
27
Correct Answer D. 27
Explanation The correct answer is 27 because when you multiply 3 by 9, you get the product of 27.
Rate this question:
14.
5×5=......
A.
26
B.
25
C.
23
D.
30
Correct Answer B. 25
Explanation The correct answer is 25 because when you multiply 5 by 5, you get the product of 25. | 677.169 | 1 |
Common Mistakes Made in ICSE Class 10 Board Exams - Chapter Circles
In this blog, we shall examine the world of circles, underlining the difficulties that frequently cause learners to give up. It tries to simplify the topic by pointing out and fixing the most prevalent mistakes made by students.
Circles have gained a lot of attention from students as well as some great Mathematicians, because of its perfect symmetry and endless applications. Despite its simplicity, navigating through the details of circles can prove challenging for students.In this blog, we'll delve into the most frequent mistakes made by students while attempting various questions based on circles. By identifying these errors and offering some insights to overcome the mistakes, we aim to encourage students to approach the questions based on circles with confidence and clarity. So, let's first get clear with the concepts of circles and then delve into the common mistakes.
Overview of Circles:
A figure obtained by joining all those points in a plane which are at a constant distance from a fixed point in the same plane is called a circle.
The fixed point is called the centre of the circle.
Constant distance is called the radius.
The line segment joining any two points on the circumference of the circle is called a chord of the circle.
The longest chord or a chord passing through the centre of a circle is the diameter.
A line that touches a circle is a tangent to the circle and the point at which the line touches the circle is a point of contact. Moreover, the length of the tangent is the distance between the external point P and the point of contact with the circle.
One of the most important parts of a circle is a cyclic quadrilateral which is a quadrilateral inscribed in a circle.
There are a few theorems based on the circles which are listed below:
The angle which, an arc of a circle subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
Angles in the same segment of a circle are equal.
The angle in a semi-circle is a right angle.
The opposite angles of a cyclic quadrilateral are supplementary.
If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
It happens that the students do not apply the correct theorem or sometimes the theorem is correct but the execution does not go well. So, it is very important to recall the theorems when solving questions based on circles.
Such mistakes have been shown below with an example below:
Solution(s) with mistake
Correct Solution
Unable to apply the theorem for length of the tangent.
OR
Applied semi-circle theorem taking AT as the diameter which is incorrect.
In short, it is very important to be thorough with the concepts and properties of circles to achieve success in mathematics. In this blog, we discussed some most common mistakes made by students and the steps that must be taken to avoid these pitfalls when dealing with circles. The key ingredients to overcome these challenges are practice, patience and a willingness to learn from mistakes. With these, you can master the fascinating world circles. So, let's embrace these lessons, keep striving for improvement, and approach circles with confidence and clarity in our mathematical journey with the help of TopperLearning. | 677.169 | 1 |
The Power of "a square + b square": Exploring the Mathematics Behind This Fundamental Equation
Mathematics is a language that allows us to understand and describe the world around us. From the simplest calculations to the most complex theories, mathematics plays a crucial role in our daily lives. One fundamental equation that has fascinated mathematicians for centuries is the expression "a square + b square." In this article, we will delve into the significance of this equation, its applications in various fields, and the insights it provides.
The Basics: Understanding "a square + b square"
Before we explore the applications and implications of "a square + b square," let's first understand what this equation represents. In simple terms, "a square + b square" is an expression that involves two variables, 'a' and 'b,' both of which are squared and then added together. Mathematically, it can be represented as:
a² + b²
This equation is a special case of the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. In this case, 'a' and 'b' represent the lengths of the two shorter sides of the triangle, and 'a square + b square' represents the square of the hypotenuse.
Applications in Geometry and Trigonometry
The equation "a square + b square" finds extensive applications in geometry and trigonometry. One of the most well-known applications is in the calculation of the distance between two points in a two-dimensional plane. This distance, often referred to as the Euclidean distance, can be calculated using the Pythagorean theorem.
For example, consider two points A(x₁, y₁) and B(x₂, y₂) in a Cartesian coordinate system. The distance between these two points can be calculated using the equation:
d = √((x₂ – x₁)² + (y₂ – y₁)²)
Here, the expression "(x₂ – x₁)² + (y₂ – y₁)²" represents "a square + b square," where 'a' is the difference in the x-coordinates and 'b' is the difference in the y-coordinates. By taking the square root of this expression, we obtain the distance between the two points.
In trigonometry, the equation "a square + b square" is used to derive the Pythagorean identities, which are fundamental relationships between the trigonometric functions sine, cosine, and tangent. These identities are widely used in various fields, including physics, engineering, and computer science.
Applications in Physics
The equation "a square + b square" has significant applications in physics, particularly in the study of forces and energy. One notable example is the calculation of the magnitude of a vector, which represents both the direction and magnitude of a physical quantity.
In physics, vectors are often represented using Cartesian coordinates. The magnitude of a vector can be calculated using the equation:
|v| = √(v₁² + v₂² + v₃²)
Here, 'v₁', 'v₂', and 'v₃' represent the components of the vector in three-dimensional space. The expression "v₁² + v₂² + v₃²" is equivalent to "a square + b square + c square," where 'a', 'b', and 'c' represent the components of the vector squared. By taking the square root of this expression, we obtain the magnitude of the vector.
Furthermore, the equation "a square + b square" is also used in the calculation of energy in various physical systems. For example, in the context of kinetic energy, the equation can be expressed as:
E = ½mv²
Here, 'm' represents the mass of an object, and 'v' represents its velocity. The expression "½mv²" is equivalent to "a square + b square," where 'a' is ½m and 'b' is v. This equation allows us to calculate the kinetic energy of an object based on its mass and velocity.
Applications in Engineering and Technology
The equation "a square + b square" finds numerous applications in engineering and technology, where it is used to solve various problems and optimize designs. One such application is in the field of signal processing, where the equation is used to calculate the magnitude of a complex number.
In signal processing, complex numbers are often represented using the rectangular form, which consists of a real part and an imaginary part. The magnitude of a complex number can be calculated using the equation:
|z| = √(Re(z)² + Im(z)²)
Here, 'Re(z)' represents the real part of the complex number, and 'Im(z)' represents the imaginary part. The expression "Re(z)² + Im(z)²" is equivalent to "a square + b square," where 'a' is Re(z) and 'b' is Im(z). By taking the square root of this expression, we obtain the magnitude of the complex number.
Additionally, the equation "a square + b square" is used in various engineering applications, such as calculating the power dissipated in electrical circuits and analyzing the stability of control systems. Its versatility and simplicity make it a valuable tool for engineers and technologists.
Conclusion
The equation "a square + b square" is a fundamental expression that has far-reaching applications in various fields, including geometry, trigonometry, physics, engineering, and technology. Whether it is used to calculate distances, magnitudes, or energies, this equation provides valuable insights and solutions to complex problems. By understanding the significance of "a square + b square," we gain a deeper appreciation for the power of mathematics in unraveling the mysteries of the universe.
Q&A
1. What is the Pythagorean theorem?
The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Mathematically, it can be represented as "a square + b square = c square," where 'a' and 'b' represent the lengths of the two shorter sides, and 'c' represents the length of the hypotenuse.
2. How is the equation "a square + b square" used in geometry?
The equation "a square + b square" is used in geometry to calculate distances between two points in a two-dimensional plane. It is also a fundamental component of the Py | 677.169 | 1 |
2.Identifytrueandfalsestatementinthefollowing:(i)Thesumofthemeasureofallanglesinanrilateralis360∘.2. Identify true and false statement in the following:(i) The sum of the measure of all angles in an rilateral is 360^{\circ} .2.Identifytrueandfalsestatementinthefollowing:(i)Thesumofthemeasureofallanglesinanrilateralis360∘.
2.Fromapointonthelinebisectorofanangleperpendicularsaredrawntothearmsoftheangle.Provethattheseperpendicularsareequalinmeasure.2. From a point on the line bisector of an angle perpendiculars are drawn to the arms of the angle. Prove that these perpendiculars are equal in measure.2.Fromapointonthelinebisectorofanangleperpendicularsaredrawntothearmsoftheangle.Provethattheseperpendicularsareequalinmeasure.
2.Provethatapointwhichisequidistantfromtheendpointsofalinesegmentisontherightbisectorofthelinesegment.2. Prove that a point which is equidistant from the end points of a line segment is on the right bisector of the line segment.2.Provethatapointwhichisequidistantfromtheendpointsofalinesegmentisontherightbisectorofthelinesegment.
3.Fillintheblankstomakethesentencestruesentences:(vi)Thesumofthemeasuresofacuteangleofarighttriangleis3. Fill in the blanks to make the sentences true sentences:(vi) The sum of the measures of acute angle of a right triangle is3.Fillintheblankstomakethesentencestruesentences:(vi)Thesumofthemeasuresofacuteangleofarighttriangleis
(iv)Howmanyacuteanglesarethereinanacuteangledtriangle?(a)1(b)2(c)3(d)notmorethan2.(iv) How many acute angles are there in an acute angled triangle?(a) 1(b) 2(c) 3(d) not more than 2 .(iv)Howmanyacuteanglesarethereinanacuteangledtriangle?(a)1(b)2(c)3(d)notmorethan2. | 677.169 | 1 |
chapter outline
Symmetry
In mathematics, symmetry is the property that divides a geometrical shape into two identical halves. Heart shape, an equilateral triangle, and a rhombus are all examples of symmetry. Such shapes are called symmetric.
Symmetry
Symmetry in Real Life
We often come across symmetry in nature. For example, the wings of a butterfly, a star fish, feathers of a peacock hives of honeybees, or snowflakes are symmetric by nature. There is also symmetry in the English alphabet.
Alphabet Line Symmetry
Several man-made objects also show symmetry like the pyramids of Egypt, the Taj Mahal and several others.
In contrast, any shape that cannot be divided into equal halves is called asymmetric which explains the difference between symmetry and asymmetry.
Asymmetry
Line of Symmetry
See the heart shape, the equilateral triangle or the English alphabets again. There is an imaginary line that divides them into halves. Such a line is called the line of symmetry. Thus, the line of symmetry is the imaginary line that divides an object into two identical halves. It is also called the axis of symmetry.
Line of Symmetry
The halves are exact mirror images. Thus, when we fold the shape along a line of symmetry, the halves overlap entirely with one another.
A shape can have one, two, or more, or even an infinite line of symmetry.
Lines of Symmetry
In mathematics, lines of symmetry are of three types: 1) vertical Line of Symmetry, 2) Horizontal Line of Symmetry, and Diagonal Line of Symmetry.
Line of Symmetry Types
Types of Symmetry
We can observe symmetry in a shape when we turn, flip, or slide it. Accordingly, there are following types of symmetry:
Types of Symmetry
Translational Symmetry
A shape shows translational symmetry when they are moved from one position to another with the same orientation in the forward and backward motion. It is sliding of an object about an axis.
Rotational Symmetry
A shape shows rotational symmetry when we rotate it around a central point at an angle other than 360°, and the outcome is the same as the shape's original appearance.
Reflectional Symmetry
A shape has a reflectional (reflective) symmetry if the line of symmetry divides the object into two equal halves such that each half is a mirror image of the other. It is thus nothing but line symmetry we studied earlier.
Glide Symmetry
It is the combination of both translation and reflection symmetry. We can alter the combination; reflection at first and translation after it, or vice-versa.
Point Symmetry
A shape has point symmetry when every part of an object has a matching part in another object but in the opposite direction.
Solved Examples
How many lines of symmetry does a pentagon have?
Solution:
A pentagon has 5 lines of symmetry.
How many lines of symmetry does a hexagon have?
Solution:
A hexagon has 6 lines of symmetry
How many lines of symmetry does the given shape have? What line of symmetry does it show?
Solution:
The given shape has one line of symmetry. It has a horizontal line of symmetry. | 677.169 | 1 |
Siap Ukk Matematika Wajib Kelas X SmType description here
2.
Upload your response
3.
Satuan derajat yang memenuhi nilai 5/24 putaran adalah ....
A.
37,50º
B.
66,67º
C.
72,50º
D.
72º
E.
75º
Correct Answer E. 75º
Explanation The correct answer is 75º because when we convert 5/24 of a full rotation into degrees, we multiply it by 360 (the number of degrees in a full rotation) and divide it by 1 (since 5/24 is already in the form of a fraction). This calculation gives us 75º.
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2
0
4.
Satuan radian yang memenuhi nilai 7/10 putaran adalah ....
A.
7/5 π rad
B.
7/10 π rad
C.
3/5 π rad
D.
1/2 π rad
E.
1/3 π rad
Correct Answer A. 7/5 π rad
Explanation The correct answer is 7/5 π rad. This is because 7/10 of a full rotation is equivalent to 7/10 of 2π radians. Simplifying this gives us 7/5 π radians.
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5.
Satuan derajat untuk ¾πrad adalah ....
A.
150º
B.
135º
C.
120º
D.
115º
E.
105º
Correct Answer B. 135º
Explanation The correct answer is 135º because when converting radians to degrees, we multiply the radian measure by 180/π. In this case, ¾π rad is equal to (3/4) * (180/π) ≈ 135º.
Explanation The value of sec ∝ is the reciprocal of cos ∝. Since sin ∝ = -9/41, we can determine the value of cos ∝ using the Pythagorean identity sin^2 ∝ + cos^2 ∝ = 1. Plugging in the given value of sin ∝, we get (-9/41)^2 + cos^2 ∝ = 1. Solving for cos ∝, we find that cos ∝ = 40/41. Therefore, the value of sec ∝ is the reciprocal of cos ∝, which is 41/40.
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7.
Letak kuadran dan nilai trigonometri dari tan 690º adalah ....
A.
K III, √3
B.
K III, √3/3
C.
K IV, -√3/3
D.
K IV, -√3
E.
K IV, 1/3√3
Correct Answer C. K IV, -√3/3
Explanation The correct answer is K IV, -√3/3.
In trigonometry, the value of the tangent function is negative in the fourth quadrant (K IV). Since the angle is 690 degrees, which is equivalent to 30 degrees past the 4th quadrant, we can use the reference angle of 30 degrees in the first quadrant. The tangent of 30 degrees is √3/3. However, since the angle is in the fourth quadrant, the value is negative. Therefore, the correct answer is K IV, -√3/3.
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8.
Jika β sudut lancip, cos β=4/5 maka tan 2β = ....
A.
24/7
B.
24/14
C.
19/14
D.
11/9
E.
11/6
Correct Answer A. 24/7
Explanation If β is an acute angle and cos β = 4/5, we can use the double angle formula for tangent to find tan 2β. The formula states that tan 2β = (2tan β)/(1-tan^2 β). Since cos β = 4/5, we can use the Pythagorean identity sin^2 β + cos^2 β = 1 to find sin β = 3/5. Using the formula, we get tan 2β = (2(3/5))/(1-(3/5)^2) = 6/4 = 3/2. Therefore, the correct answer is 24/7.
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9.
Nilai trigonometri dari tan 315º - cos 135º adalah ....
A.
-3/2 √2
B.
-1/2 √2
C.
1/2 √2
D.
√2
E.
3/2 √2
Correct Answer B. -1/2 √2
Explanation The question asks for the trigonometric value of tan 315° - cos 135°. To find this value, we can use the unit circle.
First, we find the reference angle for 315°, which is 45°. The tangent of 45° is 1.
Next, we find the reference angle for 135°, which is 45°. The cosine of 45° is 1/√2.
Substituting these values into the expression, tan 315° - cos 135°, we get 1 - 1/√2. Rationalizing the denominator, this simplifies to (√2 - 1)/√2.
Explanation The correct answer is ii, iii, iv. The formula sin2a + cos2a = 1 represents the Pythagorean identity, which states that the square of the sine of an angle plus the square of the cosine of the same angle is always equal to 1. The formula cotan2a + 1 = cosec2a represents the relationship between the cotangent and cosecant functions. And the formula sec2a - 1 = tan2a represents the relationship between the secant and tangent functions.
Nilai sin x pada fungsi trigonometri yang bernilai sama adalah ....
A.
0º, 90º, 270º
B.
0º, 180º, 360º
C.
30º, 150º, 210º
D.
45º, 135º, 315º
E.
60º, 120º, 330º
Correct Answer B. 0º, 180º, 360º
Explanation The given correct answer is 0º, 180º, 360º. This is because the sine function has a periodicity of 360º, meaning that it repeats itself every 360º. The sine function is equal to 0º at 0º, 180º, and 360º, as the sine of these angles is 0. Therefore, these angles have the same sine value.
Explanation The angles that contain the maximum and minimum values of the trigonometric function f(x) = sin x are 90º and 270º. This is because the sine function reaches its maximum value of 1 at 90º and its minimum value of -1 at 270º.
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15.
Diketahui ∆PQR mempunyai panjang PQ = 464 cm,
A.
464√3 cm
B.
464 cm
C.
332√3 cm
D.
232√2 cm
E.
232 cm
Correct Answer B. 464 cm
Explanation The given answer, 464 cm, is the correct answer because it matches the given information that the length of PQ in triangle PQR is 464 cm.
Explanation Based on the given information, we can use the Law of Cosines to find the length of AB. The Law of Cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides minus twice the product of the two sides and the cosine of the included angle. In this case, we have BC = 10 cm, AC = 16 cm, and ∠ACB = 60°. Plugging these values into the Law of Cosines, we get AB^2 = 10^2 + 16^2 - 2(10)(16)cos(60°). Simplifying this equation, we find AB^2 = 100 + 256 - 320(0.5) = 100 + 256 - 160 = 196. Taking the square root of both sides, we get AB = 14 cm. Therefore, the length of AB is 14 cm.
Explanation The given triangle ABC has two sides of equal length, AB and BC, which means it is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, since AB = BC, angle BCA is equal to angle BAC, which is given as 45º. Since the sum of all angles in a triangle is 180º, we can find the measure of angle BCA by subtracting the measures of angles BAC and BCA from 180º. Thus, angle BCA is 180º - 45º - 45º = 90º. However, the given options do not include 90º, so the closest option is 60º. | 677.169 | 1 |
Understanding Elementary Shapes Class 6 Ex 5.1
Ex 5.1 Class 6 Maths Question 1. What is the disadvantage in comparing line segment by metre observation? Solution: Comparing the lengths of two line segments simply by 'observation' may not be accurate. So we use divider to compare the length of the given line segments.
Ex 5.1 Class 6 Maths Question 2. Why is it better to use a divider than a ruler, while measuring the length of a line segment? Solution: Measuring the length of a line segment using a ruler, we may have the following errors: (i) Thickness of the ruler (ii) Angular viewing These errors can be eradicated by using the divider. So, it is better to use a divider than a ruler, while measuring the length of a line segment.
Let KL = 3.5 cm LM = 3.5 cm KM = 3.5 cm and KL + LM = 3.5 cm + 3.5 cm = 7 cm 7 cm > 3.5 cm Solution: (i) For one-fourth revolution, we have So, KL + LM > KM Hence, the sum of any two sides of a triangle is greater than the third side. Hence, we conclude that the sum of any two sides of a triangle is never less than the third side.
Ex 5.2 Class 6 Maths Question 2. Where will the hand of a clock stop if it (a) starts at 12 and makes 1/2 of a revolution, clockwise? (b) starts at 2 and makes 1/2 of a revolution, clockwise? (c) starts at 5 and makes 1/2 of a revolution, clockwise? (d) starts at 5 and makes 1/2 of a revolution, clockwise? Solution: (a) Starting from 12 and making 1/2 of a revolution, the clock hand stops at 6.
(d) Starting from 5 and making 1/2 of a revolution, the clock hand stops at 2.
Ex 5.2 Class 6 Maths Question 5. Find the number of right angles turned through by the hour hand of a clock when it goes from (a)3 to 6 (b) 2 to 8 (c) 5 to 11 (d) 10 to 1 (e) 12 to 9 (f) 12 to 6 Solution: (a) 3 to 6
Ex 5.2 Class 6 Maths Question 6. How many right angles do you make if you start facing (a) south and turn clockwise to west? (b) north and turn anticlockwise to east? (c) west and turn to west? (d) south and turn to north? Solution:
(b) Starting from 7 and turning through 2 right angles, the hour hand stops at 7.
Ex 5.4 Class 6 Maths Question 4. Measure the angles given below using the protractor and write down the measure.
∴ Measure of angle (a) = 45° and the measure of angle (b) = 60°
Ex 5.4 Class 6 Maths Question 7. Fill in the blanks with acute, obtuse, right or straight: (a) An angle whose measure is less than that of a right angle is ……… . (b) An angle whose measure is greater than that of a right angle is ……… . (c) An angle whose measure is the sum of the measures of two right angles is ……… . (d) When the sum of the measures of two angles is that of a right angle, then each one of them is ……… . (e) When the sum of the measures of two angles is that of a straight angle and if one of them is acute then the other should be ……… . Solution: (a) acute (b) obtuse (c) straight (d) acute (e) obtuse
Ex 5.4 Class 6 Maths Question 8. Find the measure of the angle shown in each figure. (First estimate with your eyes and than find the actual measure with a protractor).
Ex 5.4 Class 6 Maths Question 9. Find the angle measure between the hands of the clock in each figure:
Solution: (i) The angle between hour hand and minute hand of a clock at 9.00 a.m = 90° (ii) The angle between the hour hand and minute hand of a clock at 1.00 p.m = 30° (iii) The angle between the hour hand and minute hand of a clock at 6.00 p.m = 180°.
Ex 5.4 Class 6 Maths Question 10. Investigate: In the given figure, the angle measures 30°. Look at the same figure through a magnifying glass. Does the angle becomes larger? Does the size of the angle change?
Angle
Measure
Type
∠AOB
∠AOC
∠BOC
∠DOC
∠DOA
∠DOB
Solution:
Angle
Measure
Type
∠AOB
40°
Acute angle
∠AOC
125°
Obtuse angle
∠BOC
85°
Acute angle
∠DOC
95°
Obtuse angle
∠DOA
140°
Obtuse angle
∠DOB
180°
Straight angle
Understanding Elementary Shapes Class 6 Ex 5.5
Ex 5.5 Class 6 Maths Question 1. Which of the following are models for perpendicular lines: (a) The adjacent edges of a table top. (b) The lines of a railway track. (c) The line segments forming a letter 'L'. (d) The letter V. Solution: (a) Yes, the adjacent edges of a table top are the models of perpendicular lines. (b) No, the lines of a railway tracks are parallel to each other. So they are not a model for perpendicular lines. (c) Yes, the two line segments of'L' are the model for perpendicular lines. (d) No, the two line segments of 'V' are not a model for perpendicular lines.
Understanding Elementary Shapes Class 6 Ex 5.7
Ex 5.7 Class 6 Maths Question 1. Say True or False: (a) Each angle of a rectangle is a right angle. (b) The opposite sides of a rectangle are equal in length. (c) The diagonals of a square are perpendicular to one another. (d) All the sides of a rhombus are of equal length. (e) All the sides of a parallelogram are of equal length. (f) The opposite sides of a trapezium are parallel. Solution: (a) True (b) True (c) True (d) True (e) False (f) False
Ex 5.7 Class 6 Maths Question 2. Give reasons for the following: (a) A square can be thought of as a special rectangle. (b) A rectangle can be thought of as a special parallelogram. (c) A square can be thought of as a special rhombus. (d) Square, rectangles, parallelograms are all quadrilaterals. (e) Square is also a parallelogram. Solution: (a) A square has all the properties as that of rectangle. So, it is a special rectangle. (b) A rectangle has the same properties as that of parallelogram. So, it is a special parallelogram. (c) A square has the same properties as that of a rhombus. So, it is a special rhombus. (d) Square, rectangles and parallelogram are all quadrilateral as they are all enclosed by four sides.
Ex 5.7 Class 6 Maths Question 3. A figure is said to be regular if its sides are equal in length and angles are equal in measure. Can you identify the regular quadrilateral? Solution: Square is only the regular quadrilateral with equal sides and equal angles. Therefore, square is a regular quadrilateral.
Understanding Elementary Shapes Class 6 Ex 5.8
Ex 5.8 Class 6 Maths Question 1. Examine whether the following are polygons. If any one among them is not, say why?
(b) A Triangle Examples:
Ex 5.8 Class 6 Maths Question 3. Draw a rough sketch of a regular hexagon. Connecting any three of its vertices, draw a triangle. Identify the type of the triangle you have drawn. Solution: ABCDEF is a rough sketch of a regular hexagon. If we join any three vertices like D, A and B, we get a scalene triangle DAB.
Ex 5.8 Class 6 Maths Question 5. A diagonal is a line segment that joins any two vertices of the polygon and is not a side of the polygon. Draw a rough sketch of a pentagon and draw its diagonals. Solution: A B C D E is the rough sketch of a pentagon. By joining its any two vertices, we get, the following diagonals. | 677.169 | 1 |
How are central angles related to arcs and chords?
The degree measure of a minor arc is equal to the measure of the central angle that intercepts it. If two arcs are congruent then their corresponding chords are congruent. If two chords are congruent then their corresponding arcs are congruent.
How can you identify arcs chords central angles and inscribed angles?
A chord of a circle is a line segment whose endpoints lie on the circle. An inscribed angle is the angle formed by two chords having a common endpoint. The other endpoints define the intercepted arc. The central angle of the intercepted arc is the angle at the midpoint of the circle.
The measure of an arc refers to the arc length divided by the radius of the circle. The arc measure equals the corresponding central angle measure, in radians. That's why radians are natural: a central angle of one radian will span an arc exactly one radius long.
What is central angle of a chord?
Central angle: the angle at the center of a circle between two radii. Inscribed angle: the angle of a vertex on a circle between its chords. An angle subtended by the arc: A vertex whose chords' endpoints enclosed the given arc.
How do you write central angles?
A central angle is an angle with its vertex at the center of a circle, with its sides containing two radii of the circle. In the figure above, ∠PZQ,∠QZR , and ∠RZP are central angles. Sum of Central Angles: The sum of the measures of the central angles of a circle with no points in common is 360° .
A minor arc is the shorter arc connecting two endpoints on a circle . The measure of a minor arc is less than 180° , and equal to the measure of the arc's central angle . A major arc is the longer arc connecting two endpoints on a circle.
How do you identify and name the semicircle minor arc and major arc?
Every pair of endpoints defines two arcs. An arc whose measure is less than 180 degrees is called a minor arc. An arc whose measure is greater than 180 degrees is called a major arc. An arc whose measure equals 180 degrees is called a semicircle, since it divides the circle in two.
What are ARCs in shows?
A story arc (also narrative arc) is an extended or continuing storyline in episodic storytelling media such as television, comic books, comic strips, boardgames, video games, and films with each episode following a dramatic arc. On a television program, for example, the story would unfold over many episodes.
A central angle which is subtended by a semicircle has a measure of 180°. The "measure" associated with an arc. A circle is associated with a complete rotation, which is 360°. A semi-circle is associated with half of a rotation which is 180°.
Which arc is a major arc?
An arc whose measure is greater than 180 degrees is called a major arc. An arc whose measure equals 180 degrees is called a semicircle, since it divides the circle in two | 677.169 | 1 |
KCSE MATHEMATICS QUESTIONS AND SOLUTIONS ~ Topically Analyzed
KCSE Mathematics Questions With Answers
Form 3 Mathematics
The points P. Q and R lie on a straight line. The position vectors of P and R are 2i + 2j + 13k and 5i – 3j + 4k respectively. Q divides PR internally in the ratio 2:1
Find the
(a) Position vector of Q. | 677.169 | 1 |
COR. 1.-From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles.
COR. 2.-And consequently that all the angles made by any number of lines meeting in one point are together equal to four right angles.
PROP. XVI.-THEOREM.
If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.
Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAČ.
B
F
E
Bisect (I. 10.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G.
Because AE is equal to EC, and BE to EF;
1.
and (I. 15.)
2.
AE, EB, are equal to CE, EF, each to each;
The angle AEB is equal to the angle CEF,
because they are opposite vertical angles; therefore (I. 4.)
3. The base AB is equal to the base CF, and the triangle
AEB to the triangle CEF,
and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite: wherefore,
4.
The angle BAE is equal to the angle ECF;
but the angle ECD is greater than the angle ECF; therefore
In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG that is (I. 15.)
6. The angle ACD is greater than the angle ABC.
Therefore if one side, &c.
q.E.D.
PROP. XVII.-THEOREM.
Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles together are less than two right angles.
A
B
Produce BC to D; and because ACD is the exterior angle of the triangle ABC, (I. 16.)
1. ACD is greater than the interior and opposite angle ABC;
to each of these add the angle ACB; therefore
2. The angles ACD, ACB, are greater than the angles ABC, ACB;
ACD, ACB, are together equal to two right angles;
4. The angles ABC, BCA, are less than two right angles. In like manner, it may be demonstrated, that
5. BAC, ACB, as also CAB, ABC, are less than two right angles.
Therefore any two angles, &c. Q.E.D.
The greater side of every triangle is opposite to the greater angle.
Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC' is also greater than the angle BCA.
B
Because AC is greater than AB, make (I. 3.) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, (I. 16.)
1. ADB is greater than the interior and opposite angle ᎠᏟᏴ ;
but (I. 5.)
2. ADB is equal to ABD,
because the side AB is equal to the side AD; therefore likewise 3. The angle ABD is greater than the angle ACB,
wherefore much more
4. The angle ABC is greater than ACB.
Therefore the greater side, &c.
Q.E.D.
PROP. XIX.-THEOREM.
The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.
Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB.
B
For if it be not greater, AC must either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal (I. 5.) to the angle ACB; but it is not; therefore
1. AC is not equal to AB;
neither is it less; because then the angle ABC would be less (I. 18.) than the angle ACB; but it is not; therefore
2.
The side AC is not less than AB ;
and it has been shown that it is not equal to AB; therefore 3. AC is greater than AB.
Wherefore the greater angle, &c.
Q.E.D.
PROP. XX.-THEOREM.
Any two sides of a triangle are together greater than the third side.
Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC, greater than the side BC; and AB, BC, greater than ɅC; and BC, CA, greater than AB.
B
Produce BA to the point D, and make (I. 3.) AD equal to AC; and join DC.
Because DA is equal to AC, likewise (I. 5.)
1.
The angle ADC is equal to ACD;
but the angle BCD is greater than the angle ACD; therefore
2.
The angle BCD is greater than the angle ADC;
and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater (I. 19.) side is opposite to the greater angle; therefore
3. The side DB is greater than the side BC;
but DB is equal to B▲ and AC; therefore
4. The sides BA, AC, are greater than BC.
In the same manner it may be demonstrated that
5. The sides AB, BC, are greater than CA; and BC, CA,
are greater than AB.
Therefore any two sides, &c. Q.E.D.
PROP. XXI.-THEOREM.
If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle.
Let the two straight lines BD, CD, be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC, of the triangle, but contain an angle BDC greater than the angle BAC.
A
E
D
B
Produce BD to E; and because two sides of a triangle are greater than the third side (I. 20.)
1. The two sides BA, AE, of the triangle ABE, are greater
than BE;
to each of these add EC; therefore
2. The sides BA, AC, are greater than BE, EC.
Again, because (I. 20.)
3.
than CD,
The two sides CE, ED, of the triangle CED, are greater
add DB to each of these; therefore (Ax. 4.)
4. The sides CE, EB, are greater than CD, DB;
but it has been shown that BA, AC, are greater than BE, EC; much more then
5. BA, AC, are greater than BD, DC.
Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16.)
1.
than CED.
The exterior angle BDC of the triangle CDE is greater
For the same reason,
2.
The exterior angle CEB of the triangle ABE is greater than BAC;
and it has been demonstrated that the angle BDC is greater than CEB; much more then
3. The angle BDC is greater than the angle BAC.
Therefore, if from the ends of, &c. Q.E.D.
PROP. XXII.-PROBLEM.
To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third, (I. 20.)
Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third; viz. A and B greater than C; 4 and C greater than B; and B and C than 4. It is required to make a triangle, of which the sides shall be equal to A, B, C, each to each.
Take a straight line_DE terminated at the point D, but unlimited towards E, and make (I. 3.) DF equal to A, FG equal to B, and GH equal to C; and from the centre F, at a distance FD, describe (Post. 3.) the circle DKL; and from the centre G, at the distance GH, describe another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C. | 677.169 | 1 |
Manual of United States Surveying: System of Rectangular Surveying Employed ...
the terminating point 6 on the line F G, and move the rule, parallel, to the angle 5, finding the point D on the base E H, from which point draw a line to 6, the process then being complete. The line D 6 thus drawn, leaves the same area of lake to the left, that there is of land to the right. (Fig. 5.)
Any figure may be calculated upon the same principle
ceed in like manner with that portion of the figure to the left of the perpendicular line, throwing it into two triangles. (Fig. 6.)
CONVENIENT RULES FOR CORRECTING THE COURSE OF RANDOM LINES, WHEN THE CORRECTION DOES NOT EXCEED 200 LINKS TO EACH MILE.
RULE FOR HALF A MILE, OR FORTY CHAINS.
From the number of links to be corrected in that distance, subtract one-seventh; the difference will be the number of minutes of a degree required for the correction of the course.
Example.
Number of links to be corrected, 42-6= 36' answer.
RULE FOR ONE MILE, OR EIGHTY CHAINS.
From half of the number of links to be corrected in that distance, subtract one-seventh, the difference will be
the number of minutes of a degree required for the correction of the course.
Example.
Number of links to be corrected, 70 ÷ 2 = 35—5= 30' answer.
RULE FOR THREE MILES.
Divide the whole number of links to be corrected by seven; the quotient will be the number of minutes of a degree required for the correction of the course.
Example.
Number of links to be corrected, 297 ÷ 7 = 423' an
swer.
RULE FOR SIX MILES.
Divide one-half of the number of links to be corrected by seven; the quotient will be the number of minutes required for the correction of the course.
Example.
Number of links to be corrected, 370 ÷ 2 = 185÷7= 263' answer.*
The distances given for corrections in the above examples, are those for which corrections are generally made in the survey of the public lands, and the calculation for the course of the corrected line can generally be mentally made by the surveyor, while he is occupied in adjusting his instrument.
*The above rules are close approximations.
TABLE VI.
Showing the Difference of Latitude and Departure in running 80 chains, at any course from 1 to 60 minutes.
VARIATION OF THE NEEDLE.
1. The angle which the magnetic meridian makes with the true meridian, at any place on the surface of the earth, is called the variation of the needle at that place, and is east or west, according as the north end of the needle lies on the east or west side of the true meridian.
2. The variation is different at different places, and even at the same place it does not remain constant for any length of time. The variation is ascertained by comparing the magnetic with the true meridian.
3. If we suppose a line to be traced through those points on the surface of the earth, where the needle points directly north, such a line is called the line of no variation. At all places lying on the east of this line, the variation
of the needle is west; at all places lying on the west of it, the variation is east.
4. The public is much indebted to Professor Loomis for the valuable results of many observations and much scientific research on the dip and variation of the needle, contained in the 39th and 42d volumes of Silliman's Journal.
The variation at each place was ascertained for the year 1840; and by a comparison of previous observations and the application of known formulas, the annual motion, or change in variation, at each place, was also ascertained, and both are contained in the tables which follow.
5. If the annual motion was correctly found, and continues uniform, the variation at any subsequent period can be ascertained by simply multiplying the annual motion by the number of years, and adding the product, in the algebraic sense, to the variation in 1840. It will be observed that all variations west are designated by the plus sign; and all variations east, by the minus sign. The annual motions being all west, have all the plus sign.
6. Our first object will be to mark the line, as it was in 1840, of no variation. For this purpose we shall make a table of places lying near this line.
At the point whose latitude is 40° 53', longitude 80° 13',
the variation of the needle was nothing in the year 1840,
and the direction of the line of no variation, traced north, was N. 24° 35' west. The line of no variation, prolonged, passed a little to the east at Cleveland, in Ohio-the variation there being 19' east. Detroit lay still further to the west of this line, the variation there being 1° 56' east; and Mackinaw still further to the west, as the variation at that place was 2° 08′ east.
The course of the line of no variation, prolonged southerly, was S. 24° 35′ E. Marietta, Ohio, was west of this line the variation there being 1° 24' east. Charlottesville, in Virginia, was a little to the east of it—the variation there being 19' west; while Charleston, in South Carolina, was on the west- the variation there being 2° 44' east.
From these results, it will be easy to see about where the line of no variation is traced in our own country. 7. We shall give two additional tables: | 677.169 | 1 |
Elements of Geometry and Trigonometry
From inside the book
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Page 22 ... dicular , cutting off equal distances on the other line , will be equal . 3d , Of two oblique lines , drawn at pleasure , that which is farther from the perpendicular will be the longer . Let A be the given point , DE the given 22 ...
Page 24 ... dicular , and therefore equal ( Prop . XV . ) . So , likewise , are the two oblique lines AE , EB , the A two AF , FB , and so on . Therefore every point in the perpendicular is equally distant from the extremities A and B. E B Secondly ...
Page 46 ... dicular FG , it is also equally distant from the two points B and C : hence the three distances OA , OB , OC , are equal ; there- fore the circumference described from the centre O , with the radius OB , will pass through the three ...
Page 58 ... dicular on this line . Let A be the point , and BD the straight line . From the point A as a centre , and with a radius sufficiently great , describe an arc cutting the line BD in the two points B and D ; then mark a point E , equally ... | 677.169 | 1 |
Woolwich Mathematical Papers for Admission Into the Royal Military Academy: For 1891-1900 40.
УелЯдб 7 ... inclined to the vertical , draw a triangle representing the forces acting upon the ring . 2. A rectangular box ... plane containing the forces is equal to the moment of their resultant . OA , OB are chords , 4 and 5 inches in length , of ...
УелЯдб 10 ... plane , determine by geometrical con- struction the positions of the foci and directrices of the curve of section . Show that sections made by parallel planes ... inclined at an angle 0 to the axis , may be written in the form y = ( x - a - a ...
УелЯдб 15 ... inclined plane , if the work done in drawing a heavy body up a given length of it , is equal to that done in drawing the body along an equal length of a rough floor ( coefficient of friction ) . If the floor and plane are equally rough ...
УелЯдб 7 ... plane . A smooth wall is inclined at an angle of 60 ° to the horizon ; a heavy uniform rod AB 4 √6 feet long , is ... inclination to the horizon of the thread of the screw . 6. Find the relation between the power and weight in that ...
УелЯдб 8 ... inclined planes have a common altitude , and their inclinations are 30 ° and 60 ° to the horizon . Two masses start simul- taneously from the common vertex to fall one down each plane . Compare ( 1 ) their times of falling to the bottom ...
УелЯдб 2 - In every triangle, the square of the side subtending either of the acute angles is less than the squares of the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.
УелЯдб 1 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
УелЯдб 18УелЯдб 2 - ... the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse...
УелЯдб 17 - If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square on the other part. | 677.169 | 1 |
Geometry, Innovation, Uses, and Benefits
November 2, 2023
What is Geometry? Here are some key aspects of geometry:
1. Innovations in Geometry:
i) Euclidean Geometry: Developed by the ancient Greek mathematician Euclid, this is the most well-known and widely studied form of geometry. It deals with the properties of flat surfaces and is based on a set of axioms and postulates.
ii)Non-Euclidean Geometry: In the 19th century, mathematicians like Nikolai Lobachevsky, János Bolyai, and Carl Friedrich Gauss explored geometries that did not adhere to Euclid's parallel postulate. This led to the development of non-Euclidean geometries, including hyperbolic and elliptic geometry.
iii) Differential Geometry: This branch of geometry is concerned with the study of curves and surfaces, involving concepts like curvature and geodesics. It has applications in physics, particularly in the theory of relativity.
Topology: Topology is a branch of geometry that focuses on the properties of space that are preserved under continuous deformations. It deals with concepts like continuity, open sets, and connectedness.
Algebraic Geometry: This field combines algebra and geometry to study geometric objects defined by polynomial equations. It has applications in areas like cryptography and coding theory.
2. Uses of Geometry:
Engineering and Architecture: Geometry is crucial in the design and construction of buildings, bridges, and other structures. It ensures that structures are stable and adhere to safety standards.
Navigation: Geometry is used in navigation and cartography to determine distances, angles, and positions, especially in fields like GPS technology and aviation.
Computer Graphics: Geometry is a fundamental component of computer graphics, used to model and render three-dimensional objects and scenes in video games, animation, and simulations.
Physics: Geometry plays a vital role in physics, helping describe the behavior of particles, the curvature of spacetime in general relativity, and the shapes of objects in mechanics and fluid dynamics.
Art and Design: Artists and designers use geometric principles to create aesthetically pleasing compositions and structures in various art forms and design projects.
Astronomy: Astronomers use geometry to calculate distances, positions, and movements of celestial objects. Concepts like trigonometry and spherical geometry are vital in this field.
Robotics: Geometry is essential for robotic navigation and manipulation tasks, ensuring that robots can navigate through space and interact with their environment effectively.
3. Benefits of Geometry:
Spatial Awareness: The study of geometry enhances spatial awareness, which is valuable in everyday life, from arranging furniture to reading maps.
Scientific Advancements: Innovations in geometry have led to significant breakthroughs in various scientific disciplines, from physics to computer science.
Practical Applications: Geometry is used in a wide range of practical applications, from construction to technology, making our daily lives more efficient and safer.
Cultural and Artistic Significance: Geometry has played a significant role in art, architecture, and culture throughout history, contributing to the aesthetic and intellectual development of societies.
Conclusion:
In conclusion, geometry is a fundamental branch of mathematics with a rich history and a wide range of applications in various fields. Its innovations have shaped our understanding of the physical world and have led to practical solutions in engineering, technology, and everyday life. | 677.169 | 1 |
Question 2. If the lengths of two sides of a triangle are 7 cm and 10 cm, then what can be the length of the third side? Solution: Length of two sides of a triangle is 7 cm and 10 cm. In order to draw a triangle, the third side must be less than the sum of these two sides in 7 cm + 10.2 cm = 17.2 cm
Question 3. We know that in a triangle, the sum of lengths of any two sides is greater than the length of the third side. Is the sum of any angles of a triangle also greater than the third angle? If no, draw a rough sketch to show such a case. Solution: In a triangle, the sum of any two sides must be greater than its third side but in case of its angles. It is not necessary that the sum of any two angles be more than its third angle. Such as in this triangle sum of any two angles is less than its third angle. Such as 30° + 20° = 50° < 130° | 677.169 | 1 |
domesticresignation
Pentagon ABCDE and pentagon A'B'C'D'E' are shown on the coordinate plane below:Which two transformat...
4 months ago
Q:
Pentagon ABCDE and pentagon A'B'C'D'E' are shown on the coordinate plane below:Which two transformations are applied to pentagon ABCDE to create A'B'C'D'E'? Translated according to the rule (x, y) →(x + 8, y + 2) and reflected across the x-axis Translated according to the rule (x, y) →(x + 2, y + 8) and reflected across the y-axis Translated according to the rule (x, y) →(x + 8, y + 2) and reflected across the y-axis Translated according to the rule (x, y) →(x + 2, y + 8) and reflected across the x-axis
Accepted Solution
A:
Answer: translated according to the rule (x, y) →(x + 8, y + 2) and reflected across the x-axis
Reasoning:
1) The translation options are x + 8 or x + 2 and y + 2 or y + 8.
That means that the points are translated to the right and upward.
2) Then, you need to rotate the figure over the x-axis to translate it to the fourth quadrant.
To find the answer you can choose just one point to verify the rule.
3) Using the point D (-2,2) which is translated to D' (6, - 4) and knowing that the rotation over the x-axis keeps the x-coordinate unchanged while the y-coordinate is transformed into its negative, you can conclude that
3a) first the point was translated 8 units to the right and two units upward this is to a poin with x-coordinate -2 + 8 = 6 and y-coordinate 2 + 2 = 4
3b) second the point was reflected over the x -axis keeping the same x-coordinate x = 6 and transforming the y-coordinate into y = - 4.
So, the rule has been discovered: (x, y) →(x + 8, y + 2) and reflected across the x-axis | 677.169 | 1 |
A RAT-free (``right angle triangle-free'') set is a set of points, no three of which determine a Right Triangle. Let
be the largest integer such that a RAT-free subset of size is guaranteed to be contained in any set of
coplanar points. Then the function is bounded by | 677.169 | 1 |
Transversal of disjoint convex polygons
摘要
abstract = ".",
N2AB | 677.169 | 1 |
Find Angle MBC
25 Apr Find Angle MBC
Source: Find Angle MBC – Python HackerRank
ABC is a right triangle, 90° at B. Therefore, angle ABC = 90°. Point M is the midpoint of hypotenuse AC. You are given the lengths AB and BC. Your task is to find the angle of MBC in degrees. The first line contains the length of side AB. The second line contains the length of side BC. Lengths AB and BC are natural numbers. Output angle MBC in degrees. Note: Round the angle to the nearest integer. If the angle is 56.5000001°, then output 57°. If the angle is 56.5000000°, then output 57°. If the angle is 56.4999999°, then output 56°. | 677.169 | 1 |
What is the maximum area of triangle that can be inscribed in a circle?
equilateral triangle
isosceles triangle with base equal to 2r. d. An equilateral triangle having each of its side of length √3r. The triangle of the maximum area that can be inscribed in a circle is an equilateral triangle.
What is the area of the largest triangle that can be inscribed in a circle of radius 2cm?
= 1/2 × 2 × (2+2) sq cm. = 4 sq cm. Hope this helps you.
What is the area of the largest triangle that can be inscribed in a circle with radius 12?
What is the area of the largest square that can be inscribed in a circle of radius 10 Centimetre?
For any triangle △ABC, let s = 12 (a+b+c). Then the radius r of its inscribed circle is r=Ks=√s(s−a)(s−b)(s−c)s. Recall from geometry how to bisect an angle: use a compass centered at the vertex to draw an arc that intersects the sides of the angle at two points.
What is the area of the largest triangle that can be inscribed in a semicircle of radius 8 cm?
Area of tiangle=64 square centimetre.
What is the maximum area of a triangle inscribed in circle?
From the results of the Algebra and Trigonometry above, ABC is the largest area that can be constructed in such a circle. That is, a triangle with sides equidistant from the center of the circle. Discussion. Though it is intuitively true that an inscribed figure with sides equidistant from the center of the circle has the largest area] | 677.169 | 1 |
Find other quizzes for mathematics and more on. Web google classroom below are two parallel lines with a third line intersecting them. Web guided lesson notes and worksheet covering proportional parts of triangles when when a line is parallel to one side of a. Web parallel lines and proportional parts. Web parallel lines and the coordinate plane. Web browse parallel lines proportional parts resources on teachers pay teachers, a marketplace trusted by millions of. Web this parallel lines and proportional parts worksheet is suitable for 10th grade.
Proportional parts practice means progress boost your grades with. Web guided lesson notes and worksheet covering proportional parts of triangles when when a line is parallel to one side of a. Web browse parallel lines and proportional parts resources on teachers pay teachers, a marketplace trusted by millions of. Web this parallel lines and proportional parts worksheet is suitable for 10th degree. Set up a proportion using the triangle proportionality theorem.
Parallel Lines Geometry Worksheet Promotiontablecovers
Web showing 8 worksheets for proportional parts in triangles and parralel lines. Web browse parallel lines and proportional parts resources on teachers pay teachers, a marketplace trusted by millions of. What is the relationship betw the ratios of which show. Web guided lesson notes and worksheet covering proportional parts of triangles when when a line is parallel to one side.
Web google classroom below are two parallel lines with a third line intersecting them. In this parallel lines and proportional parts. Worksheets are 7 proportional parts in triangles and. Find other quizzes for mathematics and more on. Web this parallel lines and proportional parts worksheet is suitable for 10th grade.
Geometry 7.4 Parallel Lines and Proportional Parts YouTube
Web proportional parts in triangles and parallel lines date_____ period____ find the missing length indicated. Web parallel lines and the coordinate plane. Web guided lesson notes and worksheet covering proportional parts of triangles when when a line is parallel to one side of a. Web investigate where happens although parallel lines cut across 2 transversals. In this parallel cable and.
Proportional Parts of Parallel Lines YouTube
What is the relationship betw the ratios of which show. Web investigate where happens although parallel lines cut across 2 transversals. Worksheets are 7 proportional parts in triangles and. Set up a proportion using the triangle proportionality theorem. Web proportional parts in triangles and parallel lines date_____ period____ find the missing length indicated.
Parallel Lines And Proportional Parts Worksheet Answers —
Web showing 8 worksheets for proportional parts in triangles and parralel lines. Find other quizzes for mathematics and more on. Web browse parallel lines and proportional parts resources on teachers pay teachers, a marketplace trusted by millions of. Solve the proportion from step 1 for the unknown. Points in the coordinate plane.
W worksheet by kuta software llc geometry. Web if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. Web parallel lines and the coordinate plane. Web this parallel lines and proportional parts worksheet is suitable for 10th degree. Web browse parallel lines and proportional parts resources on teachers pay teachers, a marketplace trusted by millions of.
What is the relationship betw the ratios of which show. In this parallel lines and proportional parts. Web if three or more parallel lines intersect two transversals, then they cut off the transversals proportionally. W worksheet by kuta software llc geometry. | 677.169 | 1 |
straight lines that contain the angle meet one another , is put ' between the other two letters , and one of these two is some- ' where upon one of those straight lines , and the other upon ' the other line ... AB , CB is named the angle ABC ...
Page 13Page 15 ... angle DEF , and the angle ACB to DFE . B CE D F For , if the triangle ABC be applied to DEF , so that the point A may be on D , and the straight line AB upon DE ; the point B shall coincide with the point E , because AB is equal to DE ; and ...
Page 20 ... straight line AB is divided into two equal parts in the point D. Which was to be done . C A D B PROP . XI . PROB . To draw a straight line at right angles to a given straight line , from a given point in the same . Let AB be a given ...
Page 21 ... straight lines can- A not have a common segment . PROP . XII . PROB . E D B C C To draw a straight line perpendicular to a given straight line of an unlimited length , from a given point without it . Let AB be the given straight line | 677.169 | 1 |
Let's take an example of the cos inverse of 0 which is equal to 90 degrees. Since the cosine of 90 degrees is 0 so the cosine inverse of 0 is 90 degrees. This is an interesting way to understand how the inverse trigonometric functions compute the result.
Task: Try using the NumPy Arccos function with other inputs as well and observe the outputs. | 677.169 | 1 |
The Sectors Of A Circle
A sector is said to be a part of a circle made of the arc of the circle along with its two radii.
Start Quiz
John owns 8 black shirts, 7 red shirts, 6 blue shirts and 4 white shirts. If he wants to make a circle chart of his shirts, what is the degree angle corresponding to the "blue shirt section?"
The answer cannot be determined from the above information.25.0 degrees90 degrees86.4 degrees43.2 degrees
Quantity A: The angle of a circle"s sector having an arc length of 10π and a radius of 7.
Quantity B: The angle of a circle"s sector having an area of 24π and a radius of 9.
Which of the following relations is true?
The relationship between the two values cannot be determined.Quantity B is larger than quantity A.The two quantities are equal.Quantity A is larger than quantity B.
What is the perimeter of a pie piece if the pie is sliced into 40 degree pieces and its area is 361π?
38π/938 + 38π/92.138π40.1π
What is the length of the arc of a circle with radius 10 that traces a 50 degree angle?
25π/925π/2925π/725π20π/9
An ant walks around the edge of circular pizzas left on the counter of a pizza shop. On most days, it is shaken off the pizza before it manages to walk the complete distance.
Quantity A: The distance covered by the ant when walking over four slices of a pizza with a diameter of 12 and 10 equally-sized pieces.
Quantity B: The distance covered by the ant when walking over a complete personal pizza with a diameter of 6 inches.
Quantity A is larger.The relationship between the two quantities cannot be determined.Quantity B is larger.The two quantities are equal.
An ant begins at the center of a pie with a 12" radius. Walking out to the edge of pie, it then proceeds along the outer edge for a certain distance. At a certain point, it turns back toward the center of the pie and returns to the center point. Its whole trek was 55.3 inches. What is the approximate size of the angle through which it traveled?
81.53°128.21°149.52°91.44°74.76°
If the outer arc of 1/12th of a circular pie is 7π, what is the area of 1/4th of the pie?
None of the other answers21π1764π441π10.5π
What is the area of a sector of a circle of radius 4 that spans an arc of 86 degrees?
1/5 * π172/45 * π12/53 * π1/4 * π17/42 * π
A circular pie is cut into 30 pieces. Two people wish to split a piece of the pie, but one person wants to have twice as much as the other person. What is the angle of the smaller piece produced in this manner?
12 degree
8 degree
15 degree
4 degree
16 degree
A given pizza with a 10-inch diameter has 1450 calories. A baker cuts the pizza using a 25∘ angle for each piece. If Susan eats five such pieces, how many calories does she consume? Round to the nearest calorie. | 677.169 | 1 |
Which method is used in intersection of solids?
Line method: A number of lines are drawn on the lateral surface of one of the solids and in the region of the line of intersection. Points of intersection of these lines with the surface of the other solid are then located.
What is the intersection of a solid and a plane?
The intersection of a solid and a plane is called a cross section of the solid. Describe the shape resulting from a vertical, angled, and horizontal cross section of a cylinder. The cross section is a rectangle.
What is interpenetration explain with sketch?
When two soids interpenetrate, (intersect), eachother a straight line or a curved line of interpenetration is formed. Whether the line is curved or straight depends on the shape of the solids. Flat — Flat = Straight.
What is the intersection of two copies?
Step-by-step explanation: The intersection of two sets A and B, denoted by A ∩ B, is the set of all objects that are members of both the sets A and B. In symbols, That is, x is an element of the intersection A ∩ B, if and only if x is both an element of A and an element of B.
What is the cross section of a cylinder?
The cross-sectional area of a cylinder is equal to the area of a circle if cut parallel to the circular base. The cross-sectional area is the area of a two-dimensional shape that is obtained when a three-dimensional object – such as a cylinder – is sliced perpendicular to some specified axis at a point.
What is the cross section formed by the intersection of plane and a solid?
Describe the cross section formed by the intersection of the plane and the solid….
Term
Definition
cross section
A cross section is the intersection of a three-dimensional solid with a plane.
What is need of intersection of solids?
PURPOSE OF DRAWING THESE CURVES:- WHEN TWO OBJECTS ARE TO BE JOINED TOGATHER, MAXIMUM SURFACE CONTACT BETWEEN BOTH BECOMES A BASIC REQUIREMENT FOR STRONGEST & LEAK-PROOF JOINT. Curves of Intersections being common to both Intersecting solids, shows exact & maximum surface contact of both solids.
When a line intersects a circle how many points generated?
Tangent: A tangent to a circle is a line that intersects the circle at only one point. The point where the tangent intersects the circle is called the point of tangency.
When a prism intersects with the prism then we get?
When two solids bounded by plane surfaces such as prism and pyramid penetrate each other, we obtain straight lines as their line of intersection.
What happens at the intersection of two solids?
WHEN ONE SOLID PENETRATES ANOTHER SOLID THEN THEIR SURFACES INTERSECT AND AT THE JUNCTION OF INTERSECTION A TYPICAL CURVE IS FORMED, WHICH REMAINS COMMON TO BOTH SOLIDS. THIS CURVE IS CALLED CURVE OF INTERSECTION or INTERPENETRSATION (COI ) AND IT IS A RESULT OF INTERPENETRATION OF SOLIDS.
When do you need to know the shape of an intersection?
Intersection of regular solids. When two solids Interpenetrate, a line of intersection is formed. H is sometimes necessary to know the exact shape of this line, usually so that an accurate development of either or both of the solids can be drawn.
How is the line of intersection between two surfaces obtained?
The line of intersection between two plane surfaces is obtained by locating the positions of points at which the edges of one surface intersect the other surface and then joining the points by a straight line.These points are called vertices.
When do you draw a line of intersection?
When two solids Interpenetrate, a line of intersection is formed. H is sometimes necessary to know the exact shape of this line, usually so that an accurate development of either or both of the solids can be drawn. This chapter shows the lines of intersection formed when some of the simpler geometric solids interpenetrate | 677.169 | 1 |
Math Grade 8 Quiz Similarity of Two-Dimensional Figures – 8.G.A.4
Two-dimensional figures are said to be similar if one can be transformed into the other through a sequence of rotations, reflections, translations, and dilations. These transformations maintain the shape's proportionality and angle measures. When two figures are similar, their corresponding angles are congruent, and their corresponding side lengths are proportional. Understanding the concept of similarity is crucial for various applications, such as real-world modeling and geometric proofs.
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Total Number of Triangles Puzzle
Solution to this GeometryTrianglePuzzle problem is given in the video below!
Total Number of Rectangles Puzzle
How many rectangles are there in the figure below?
Solution to this GeometryRectanglePuzzleis given in the video below!
Total Number of Upright SQUARES in a Square
How many upright Squares are there in the following square?
The key is to understand vertices of any given upright square and a diagonal line of the largest square to see how many translations of that given upright square you can have. As such, there is a summation-derived FORMULA you can actually use for any largest size square to find the total number of possible upright squares contained within it!
Solution to this GeometrySquarePuzzleis given in the video below!
Total Number of Upright RECTANGLES in a Square (or Rectangle)
How many upright Rectangles are there in the following square?
The key is to understand vertices of any given upright rectangle to see how many unique translations of that rectangle are possible. As such, there is actually a summation-derived FORMULA you can utilize for the largest square or rectangle of any size to find the total number of possible upright rectangles within it!
Solution to this GeometryRectanglePuzzleis given in the video below!
Total Number of Upright SQUARES in a Rectangle
How many upright Squares are there in the following rectangle?
The key is to understand vertices of any given upright square to see how many translations of that given square you can have. As such, there is a summation-related FORMULA you can actually derive for any size rectangle to find the total number of possible upright squares contained within it!
Solution to this SquareRectangleGeometryPuzzleis given in the video below!
Total Number of Upright & Tilted SQUARES in a Square
How many upright and tiltedSquares are there in the following square?
The key is to use vertices of any given largest square to see how many translations of a given smaller square you can have. As such, there is a summation-derived FORMULA you can actually use for any size square to find the total number of possible upright and tilted squares contained within it!
Solution to this SquareGeometryPuzzle example is given in the video below!
Total Number of Upright & Tilted SQUARES in a Rectangle
How many upright and tiltedSquares are there in the following rectangle?
The key is to use vertices of any given rectangle to see how many translations of a given square you can have. As such, there is a summation-derived FORMULA you can actually use for any size rectangle to find the total number of possible upright and tilted squares contained within it!
Solution to this RectangleGeometryPuzzle example problem is given in the video below!
Ferris Wheel trigonometry word problem
The Ferris wheel at Navy Pier has a diameter of 140 feet. It stands 10 feet off the ground. The wheel has 40 gondolas that seat six passengers each. It takes about 6 minutes for the Navi Pier Ferris wheel to complete one rotation.
Draw a diagram of the Navy Pier Ferris wheel and the boarding platform. Fill in the necessary information. Sketch the graph. Write a cosine equation for your curve. Write a sine equation for your curve.
Answer the following questions:
i. What is the circumference of the wheel?
ii. At what speed is the wheel traveling? Please answer in feet / second.
iii. If you begin your ride at the base of the wheel, what is the height after 1 minute? 4 minutes?
iv. At what approximate time(s) will you reach the following heights?
a) 100 ft
b) 240 ft
v. What is the length of the arc traveled by the Navy Pier Ferris wheel from the 4 o'clock to the 7 o'clock position?
Solution to this Trigonometric Function word practice problem is provided in the video below!
Roller Coaster trigonometry problem
A portion of a roller coaster is to be built in the shape of a sinusoid. You have been hired to calculate the lengths of the horizontal and vertical timber supports to be used.
a. The high and low points on the track are separated by 50 meters horizontally and 30 meters vertically. The low point is 3 meters below the ground. Letting y be the number of meters the track is above the ground and x the number of meters horizontally from the high point, write an equation expressing y in terms of x.
b. How long is the vertical timber at the high point? At x = 4 m? At x = 32 m?
c. Where does the track first go below ground?
Solution to this Trigonometric Function example practice problem is provided in the video below!
Steamboat trigonometry example word problem
Mark Twain sat on the deck of a river steamboat. As the paddlewheel turned, a point on the paddle blade moved in such a way that its distance, d, from the water's surface was a sinusoidal function of time. When his stopwatch read 4 seconds, the point was at its highest, 16 feet above the water's surface. The wheel's diameter was 18 feet, and it completed a revolution every 10 seconds.
a. Sketch a graph of the sinusoid
b. Write the equation of the sinusoid
c. How far above the surface was the point when Mark's stopwatch read:
i. 5 seconds
ii. 17 seconds
d. What is the first positive value of time at which the point was at the water's surface? At that time, was it going into or coming out of the water? Explain.
Solution to this Trigonometric Function example word problem is provided in the video below!
Temperature trigonometry word problem
The max temperature in Buenos Aires is on January 15 and is 33 degrees Celsius. The minimum temperature is on July 16 (day 197) and is 9 degrees Celsius. (Assume the period is 365 days).
a) Sketch the temperature as a function of time
b) Find the equation for the temperature, T, as a function of time, t.
c) What is the temperature on Mother's Day, May 10?
d) Give the dates during a one year period when the temperature is below 18 degrees Celsius.
Solution to this Trigonometric Function example word problem is provided in the video below! | 677.169 | 1 |
Coordinate in Space Class 11 Mathematics Solutions | Exercise - 11.2
Sanjeev
Jun 28, 2023
Chapter - 11 Coordinate in Space
In class 11 mathematics, one of the things we learn is coordinate geometry in space. It's all about using numbers to describe points, lines, and shapes in three-dimensional space. It might sound complicated at first, but it's actually a useful tool for solving problems and understanding how things are positioned in space.
With coordinate geometry, we can easily represent and analyze objects in three dimensions, which helps us make sense of the world around us. So, even though it may seem challenging, studying coordinate geometry in space can be quite beneficial and interesting.
Many of you have been reaching out to us through our social media handles, and asking us for the solution of this chapter. We understand your need for comprehensive study materials. So, today I am here to fulfill your demand for the complete notes of the class 11 coordinate in space chapter.
Before we move towards the solution of this chapter. Let's have a short discussion on the topics we have to study in coordinate in space second exercise. We already have discussed what we have to study in first exercise in our previous article. If you haven't check that out then you should have a look at that before you start second exercise.
Projection
Projections are the transformation of points and lines on one plane onto another plane by connecting corresponding points on the two planes with parallel lines that go across those planes
a) Projection of a point on a line or a plane
The projection of a point P on a line AB or a plane CDEF is the foot L of the perpendicular from the point P on the line AB or plane CDEF. In the first case, it is also interpreted as the A- point L of intersection of the plane through the given point P and perpendicular to the given line.
b) Projection of a line segment on a line
The projection of the line segment AB of a given line on another line CD is the segment A'B' of CD where A' and B' are the projections of A and B on CD. Here we may also say that A' and B' are the points in which planes through A and B perpendicular to CD meet the line CD
Exercise - 11.2
In this PDF, you'll only find the solution of class 11 coordinate in space chapter. It contains all the solutions of exercise - 11.2. If you want the solutions of other exercises then you'll find them above. Just click on the button and you'll reach your destination.
Note: Scroll the PDF to view all Solution
You are not allowed to post this PDF in any website or social platform without permission.
How to Download Class 11 Coordinate in Space coordinate in space | 677.169 | 1 |
Unit 8 Polygons And Quadrilaterals Test Answer Key
Introduction
If you are a student who has recently taken a Unit 8 Polygons and Quadrilaterals test, then you might be eagerly waiting for the answer key. An answer key is a document that contains the correct answers to all the questions asked in the test. It helps students to evaluate their performance and identify their strengths and weaknesses. In this article, we will provide you with the answer key for the Unit 8 Polygons and Quadrilaterals test.
What are Polygons and Quadrilaterals?
Polygons and quadrilaterals are two-dimensional figures that are commonly studied in geometry. A polygon is a closed figure with three or more straight sides, while a quadrilateral is a polygon with four sides. Some examples of polygons include triangles, rectangles, pentagons, and hexagons. Some examples of quadrilaterals include squares, rectangles, rhombuses, and trapezoids.
Test Overview
The Unit 8 Polygons and Quadrilaterals test is designed to evaluate your understanding of various concepts related to polygons and quadrilaterals. The test might include questions on topics such as identifying polygons and quadrilaterals, calculating angles and sides, and finding the perimeter and area of polygons and quadrilaterals.
Answer Key
Explanation of Answers
1. The sum of the interior angles of a triangle is 180 degrees. 2. A quadrilateral is a polygon with four sides. 3. A rectangle has four right angles. 4. The opposite sides of a parallelogram are parallel and congruent. 5. The area of a triangle is calculated using the formula 1/2 * base * height. 6. A square is a special type of rectangle where all the sides are equal. 7. A rhombus has opposite sides parallel and congruent, and all four sides are equal. 8. A trapezoid is a quadrilateral with one pair of parallel sides. 9. The perimeter of a polygon is the sum of the lengths of all its sides. 10. The area of a rectangle is calculated using the formula length * width.
Conclusion
In conclusion, the Unit 8 Polygons and Quadrilaterals test is an important assessment tool for students studying geometry. The answer key provided in this article can help you evaluate your performance and identify areas where you need to improve. We hope this article has been helpful to you. Good luck with your studies! | 677.169 | 1 |
Foundations of Geometry (Vocab)
Geometry is a branch of mathematics concerned with the properties, position, measurement, and relationships of points, lines, angles, planes, and solids.
Point – A location in space. A point has no dimensions – no length, no width, no height. It only has position. We represent a point as a dot and usually name it with a capital letter such as point P.
Line – An infinite set of points with no endpoints. A straight line extends endlessly in two opposite directions. A line has only one dimension, length. We can name a line by any two points on it. For example, if two points are A and B, we have line AB or AB. We can also name a line by a single lowercase letter, line l.
Plane – A set of points forming a flat surface that extends infinitely in all directions. A plane has two dimensions, length and width. It has no thickness.
Collinear Points – A set of points all lying on the same line. ( noncollinear points do not all lie on the same line.) Any two points are collinear.
Coplanar points – A set of points all lying in the same plane. (Noncoplanar points do not all lie in the same plane) Any three points are coplanar.
Ray – Part of a line, consisting of one endpoint and all the points on one side of that endpoint. A ray is named by the endpoint and any other point. We can refer to the figure as ray AB or AB.
Line segment – Part of a line, consisting of two endpoints and all the points on the line between them. A line segment is named by the two endpoints. If the endpoints are A and B, we have line segment AB or AB or BA. The measure of the length of the segment is the distance between A and B.
Angle – (An angle is formed by two rays that share an endpoint. The rays are called sides and the endpoint is called the vertex. We can also say that angles are formed by intersecting lines – lines that cross or meet. In this case the vertex is the point of intersection.
-There are several ways to name an angle:
1) By a capital letter that names its vertex, such as
2) By a lowercase letter or a number placed inside the angle, such as
3) By three capital letters, such as . The middle letter is the vertex, and the other two letters name points on different rays (or lines), in either order. can also be named .
-Angles are measured by degrees and are classified by their measure.
-Congruent angles are equal in measure.
Classification of Angles:
1) Right angle – An angle of exactly 90 degrees. The symbol of a right angle is a square at the vertex.
2) Acute angle – An angle less than 90 degrees.
3) Obtuse angle – An angle greater than 90 degrees but less than 180 degrees.
4) Straight angle – an angle of exactly 180 degrees. A straight angle is a line.
5) Reflexive angle - an angle whose measure is greater than 180˚ and less than 360˚.
Adjacent angles – Two angles having the same vertex and sharing one side, but do not have any interior points in common.
<ABD and <DBC are adjacent
C
D
AB
Angle bisector – a ray (or any part of a line) whose endpoint is the vertex of the angle and divides the angle into two congruent (equal) angles.
<ABD = <DBC
A
D
B C
Complementary angles - two angles whose sum measures 90 degrees.
Supplementary angles - two angles whose sum measures 180 degrees.
(Two angles that create a straight line - a linear pair)
Perpendicular lines ( - two lines (or parts of lines) that intersect to form right angles. Lines with negative reciprocal slopes are perpendicular.
Vertical angles - angles formed by intersecting lines; they are opposite each other and are congruent | 677.169 | 1 |
6. Classify the triangle according to sides, that is, equilateral, isosceles and scalene triangles
(a) 6 cm, 3 cm, 5 cm.
(b) 6 cm, 6 cm, 6 cm.
(c) 7 cm, 7 cm, 5 cm.
(d) 8 cm, 12 cm, 10 cm.
7. Draw any triangle and name it PQR. Measure its sides and classify the triangle as Isosceles triangle, scalene triangle or equilateral triangle.
8. The perimeter of a triangle is 24 cm. Two of its sides are 8 cm and 9 cm. Find the length of its third side.
9. What is a triangle?
10. Find the missing side in the diagram below:
What have we learnt:
Understand Triangle and its parts
Types of triangles
Triangle and its properties
The sum of all internal angles of a triangle is always equal to 180°
The sum of the length of any two sides of a triangle is greater than the length of the third side
Understand Equilateral, Isosceles and Scalene | 677.169 | 1 |
The Elements of Plane and Solid Geometry: With Numerous Exercises
736. The area of a spherical triangle is equal to its spherical excess.
Hyp. Let A, B, C denote the A numerical measures of the s of the spherical ▲ ABC, the rt.
being
the unit of s, and the tri-rectangular A the unit of areas.
To prove area ABC=A+B+C-2. Proof. Continue any one side as AB so as to complete the great O
B
B
ABA'B'. Continue the other two sides AC and BC till they meet this O in A' and B'.
Then area ABC÷area A'BC=lune ABA'C=2A,
and area ABC+area AB'C=lune AB'CB=2B. (735)
Also, since the As ABC and A'B'C are together equivalent to a lune whose angle is C,
... area ABC + area A'B'C = 2C.
(729)
But the sum of the areas of the As ABC, A'BC, AB'C, A'B'C is equal to the area of the hemisphere, or 4.
.*. adding these three equations, we have
2 area ABC + 4 = 2A + 2B + 2C.
... area ABC=A+B+C - 2. Q.E.D.
737. COR. 1. By a method similar to that in (736), in connection with (730) and (735), it may be proved that The volume of a triangular spherical pyramid is equal to the spherical excess of its base, (the volume of the trirectangular pyramid being the unit of volume).
738. COR. 2. If the three vertices of a triangle are on a great circle, its three sides must coincide with that circle, and each angle must equal 180°. The area of the triangle is then a hemisphere, and its spherical excess 4 rt. /s, or 360°. Therefore the area of the surface of the whole sphere is 720°. Hence:
The area of a spherical triangle is to that of the surface of the sphere as its spherical excess, in degrees, is to 720°.
Proposition 20. Theorem.
739. The area of a spherical polygon is equal to its spherical excess.
Proof. From any vertex, as A, draw diagonals, dividing the polygon into (n-2) As.
Then, since the area of each ▲ the sum of its s minus 2 rt.s(736); and since the sum of the s of the (n − 2) ▲ s = the sum of the s of the polygon, or S,
... K = S − 2(n − 2).
Q. E.D.
NOTE. In the last three propositions, only the ratios of the areas are expressed. If the absolute area is required, the area of the surface of the sphere must be known.
EXERCISES.
THEOREMS.
1. The intersection of the surfaces of two spheres is a circle whose plane is at right angles to the line joining the centres of the spheres, and whose centre is in that line.
2. If lines be drawn from any point of the surface of a sphere to the ends of a diameter, they will form with each other a right angle.
3. If any number of lines in space pass through a point, the feet of the perpendiculars upon these lines from another point lie upon the surface of a sphere.
4. If two straight lines are tangent to a sphere at the same point, the plane of these lines is tangent to the sphere. 5. On spheres of different radii, mutually equiangular triangles are similar.
See (721), note.
6. The sum of the two arcs of great circles drawn from the extremities of one side of a spherical triangle to a point within it, is less than the sum of the other two sides.
7. If from any point on the surface of a sphere two arcs of great circles are drawn perpendicular to a circumference, the shorter of the two arcs is the shortest arc that can be drawn from the given point to the circumference.
8. Any lune is to a tri-rectangular triangle as its angle is to half a right angle.
9. Spherical polygons are to each other as their spherical
excesses.
10. Two oblique arcs drawn from the same point to points of the circumference equally distant from the foot of the perpendicular are equal.
11. Of two oblique arcs, the one which meets the circumference at the greater distance from the foot of the perpendicular is the longer.
12. The arc of a great circle, tangent to a small circle, is perpendicular to the radius of the sphere at the point of contact.
13. If three spheres intersect one another, their planes of intersection intersect in a right line perpendicular to the plane containing the centres of the spheres.
14. Prove that this line is the locus of points from which tangent lines to the three spheres are equal.
15. Through a fixed point, within or without a sphere, three lines mutually at right angles intersect the sphere: prove that the sum of the squares of the three chords is constant, depending only on the radius of the sphere and the distance of the point from the centre.
16. Prove also that the sum of the squares of the six segments is constant.
17. Prove that the area of a spherical triangle, each of whose angles is of a right angle, is equal to the surface of a great circle.
18. If through a point O any secant OPP' is drawn to cut a sphere in P, P', prove that OP. OP' is constant.
19. Find the radii of the spheres inscribed and circumscribed to a regular tetraedron.
NUMERICAL EXERCISES.
20. If the sides of a spherical triangle are respectively 65°, 112°, and 85°, how many degrees are there in each angle of its polar triangle?
21. If the angles of a spherical triangle are respectively 90°, 115°, and 70°, how many degrees are there in each side of its polar triangle?
22. Given the spherical triangle whose sides are respectively 80°, 90°, and 140°, to find the angles of its polar triangle.
23. What part of the surface of a sphere is a lune whose angle is 45° 54°? 80°?
24. What part of the volume of a sphere is an ungula whose angle is 72° ? 36°?
25. If the angle of a lune is 50°, find its area on a sphere whose surface is 72 square inches. Ans. 10 square inches.
26. Find the area of a spherical triangle whose angles are respectively 75°, 100°, and 115°, on a sphere whose surface is 72 square inches. Ans. 11 square inches. 27. Find the area of a spherical triangle each of whose angles is 70°, on a sphere whose surface is 144 square inches. Ans. 6 square inches. 28. Find the area of a spherical triangle whose angles are 60°, 90°, and 120°, on a sphere whose surface is 64 square inches. Ans. 8 square inches. 29. Find the area of a spherical polygon of six sides each of whose angles is 150°, on a sphere whose surface is 100 square inches. Ans. 25 square inches.
30. Find the area of a bi-rectangular triangle whose vertical angle is 108°, on a sphere whose surface is 100 square inches. Ans. 15 square inches. 31. Find the area of a spherical pentagon whose angles are respectively 138°, 112°, 131°, 168°, and 153°, on a sphere whose surface is 40 square feet. Ans. 9 square feet. 32. Find the area of a spherical triangle whose angles are 61°, 109°, and 127°, on a sphere whose surface is 10 square inches. Ans. 1.625 square inches.
33. Find the area of a spherical quadrangle whose angles are 170°, 139°, 126°, and 141°, on a sphere whose surface is 400 square inches.
34. Find the area of a spherical pentagon whose angles are 122°, 128°, 131°, 160°, and 161°, on a sphere whose surface is 150 square feet.
35. Find the angles of an equilateral spherical triangle whose area is equal to that of a great circle. Ans. 120°.
36. Find the angles of an equilateral spherical triangle whose area is equal to that of an equilateral spherical hexagon, each of whose angles is 150°. Ans. 80°. | 677.169 | 1 |
A Fundamental Theorem of Similarity
A Fundamental Theorem of Similarity
If F and F′ are any two directly similar1 figures with the vertices P in F corresponding to vertices P′ in F′, and the lines PP′ are divided in the ratio of r : 1−r, that is, at points P′′ = (1 − r)P + rP′, then the new figure F′′ formed by the points P′′ is directly similar to F and F′. Illustration: The dynamic figure below illustrates the theorem for two directly similar quadrilaterals IJKL and I′J′K′L′. Drag any of the vertices of these two directly similar quadrilaterals, or B to vary the ratio r above, or E or H to change the position or orientation of I′J′K′L′. Click on the 'Show Ratio Measurements' button to display the ratios of corresponding sides of F′′ and F′.
(Note1: Two similar figures are 'directly similar' if their corresponding angles have the same rotational sense (and are not reversed in relation to each other as in a reflection). | 677.169 | 1 |
Monday, October 17, 2022
Issue
I am trying to numerically find the solution to a problem of finding the centre and radius of a circle that is tangential to a given line and passes through another given point. The problem is shown in the following image:
The circle must be tangential to the line Y=mx + c and must pass through the point. The centre of the circle will always lie on the x-axis (y=0) which is the bisector of the two lines as the problem is symmetric. Point A is always known, as is the angle theta. Point P is specified as the input
I can write down some of the equation for the right-angle triangle: Sin theta = BC / AC and I know that BC == PC == R, but I'm not sure how to solve this numerically as the equation of the line and the point will be different every time the code is called. Solving by hand would involve substituting A and P back into the equations of the lines I think.
I presume I need to solve some simultaneous equations using some scipy module, but don't know what exactly I should do. Hopefully someone can help.
Solution
I think there are two solutions which can work out by solving a quadratic equation. | 677.169 | 1 |
The rotary movement
Two arbitrarily placed congruentequilateral triangles. Out of all the possible rotations of the plane, for how many will the image of ABC be A'B'C'?
Scroll down for a solution to this problem.
Solution
There is exactly one rotation, unless the corresponding sides are parallel, in which case there are none.
The first step is to find the location of the rotation centre by intersecting the perpendicular bisectors of AA' and BB'. If AB and A'B' are parallel, so will the bisectors be and they will not intersect. In this case there exists no rotation.
In all other cases there will be one possible rotation centre O. Now we have to prove that this O is also equidistant from C and C'. Looking at the figure above, we deduce from SSS that triangles ABO and A'B'O are congruent. Therefore angles OAC and OA'C' are equal. From SAS it follows that triangles AOC and A'OC' are congruent and hence OC and OC' are congruent.
Finally, we must have that the rotation angles of the three vertices are equal. From the figure above we see that angle AOA' equals α+β and so is angle BOB'. Therefore the are equal. Similarly they are equal to COC'.
Poem
Two twin triangles Court hopefully each other Every one does what must be done And revolves around both Successive rotations So that everyone follows His own desire A reciprocal attraction A very strong alchemy Like the two poles of a magnet Two poles of love
One reply on "The rotary movement"
1. If the corresponding sides are parallel, two rotations map ABC onto A'B'C':
For example take an arbitrary point P and map ABC by point reflection at P (that's a rotation with 180°) onto A"B"C"; then take the midpoint Q of A" and A': point reflection at Q maps A"B"C" onto A'B'C'.
2. The argumentation for the general case is shorter, if we use the centres of the circumcircles of the triangles:
Let M resp. M' be the center of ABC resp. A'B'C', gM the perpendicular bisector of M and M'.
Then for any point R of gM the rotation with centre R and angle a(M, R, M') maps M onto M' and the circumcircle of ABC onto that oh A'B'C'.
If R is the point of intersection of gM and the perpendicular bisector of A and A' then the rotation also maps A onto A'.
As rotation preserves angles, B is mapped onto B' and C onto C'. | 677.169 | 1 |
The Diagonal of a Rectangle: Exploring its Properties and Applications
A rectangle is a fundamental geometric shape that is widely used in various fields, from architecture and engineering to mathematics and design. One of the key characteristics of a rectangle is its diagonal, which connects opposite corners of the shape. In this article, we will delve into the properties of the diagonal of a rectangle, its mathematical significance, and its practical applications.
Understanding the Diagonal of a Rectangle
Before we explore the properties of the diagonal, let's first define what a rectangle is. A rectangle is a quadrilateral with four right angles, where opposite sides are equal in length. The diagonal of a rectangle is the line segment that connects two non-adjacent vertices, forming a straight line across the shape.
The In the case of a rectangle, the diagonal acts as the hypotenuse, while the sides of the rectangle act as the other two sides of the triangle.
Let's consider a rectangle with side lengths of a and b. The diagonal, denoted as d, can be calculated using the formula:
d = √(a2 + b2)
Properties of the Diagonal
The diagonal of a rectangle possesses several interesting properties that make it a valuable concept in geometry and beyond. Let's explore some of these properties:
1. Length and Proportions
The length of the diagonal is always greater than the length of any side of the rectangle. This can be proven usingFurthermore, the diagonal divides the rectangle into two congruent right-angled triangles. This means that the ratio of the length of the diagonal to the length of either side of the rectangle is equal to √2. In other words, d/a = d/b = √2.
2. Bisects the Rectangle
The diagonal of a rectangle bisects the shape into two congruent right-angled triangles. This property is useful in various applications, such as cutting materials or dividing a space into equal parts. By drawing a diagonal across a rectangle, we can easily divide it into two equal halves.
3. Maximum Distance
The diagonal of a rectangle represents the maximum distance between any two points within the shape. This property is particularly relevant in fields such as computer graphics and optimization, where finding the longest distance between two points is essential.
4. Relationship with Other Geometric Shapes
The diagonal of a rectangle is related to other geometric shapes and concepts. For example, the diagonal of a square is equal to the side length multiplied by √2. Additionally, the diagonal of a rectangle is also the hypotenuse of the right-angled triangles formed by the rectangle's sides.
Applications of the Diagonal of a Rectangle
The diagonal of a rectangle finds practical applications in various fields. Let's explore some of these applications:
1. Architecture and Construction
In architecture and construction, rectangles are commonly used for designing buildings, rooms, and structures. The diagonal of a rectangle helps architects and engineers determine the maximum distance between two corners of a room or the longest span of a beam. This information is crucial for ensuring structural stability and optimizing space utilization.
2. Design and Layout
The diagonal of a rectangle is often used in design and layout applications to create visually appealing compositions. By aligning elements along the diagonal, designers can achieve a sense of balance and harmony in their designs. This technique is commonly employed in photography, graphic design, and interior design.
3. Mathematics and Geometry
The diagonal of a rectangle is a fundamental concept in geometry and mathematics. It serves as a basis for understanding the properties of other shapes, such as squares and parallelograms. Additionally, the diagonal of a rectangle is used in various mathematical proofs and calculations.
4. Computer Graphics and Visualization
In computer graphics and visualization, the diagonal of a rectangle is essential for rendering and transforming 2D and 3D objects. It helps determine the distance between points, the orientation of objects, and the scaling of shapes. By utilizing the properties of the diagonal, computer graphics algorithms can create realistic and visually appealing images.
Q&A
1. Can the diagonal of a rectangle be longer than the perimeter?
No, the diagonal of a rectangle cannot be longer than the perimeter. The perimeter of a rectangle is equal to the sum of all its sides, while the diagonal is the longest side of the right-angled triangle formed by the rectangle. Therefore, the perimeter will always be greater than or equal to the diagonal.
2. How can the diagonal of a rectangle be used to find the area?
The diagonal of a rectangle alone cannot be used to find the area. However, by knowing the length of the diagonal and one side of the rectangle, you can use the Pythagorean theorem to calculate the length of the other side. Once you have both side lengths, you can multiply them to find the area of the rectangle.
3. Is the diagonal of a rectangle always longer than its sides?
Yes, the diagonal of a rectangle is always longer than its sides. This is a result of4. Can a rectangle have multiple diagonals?
No, a rectangle can only have one diagonal. The diagonal connects two non-adjacent vertices of the rectangle, forming a straight line across the shape. Since a rectangle has four vertices, there are only two non-adjacent pairs, and therefore, only one diagonal.
5. How does the diagonal of a rectangle compare to the diagonal of a square? | 677.169 | 1 |
Construction of regular polygons
Regular pentagon
The problem of constructing a regular pentagon is equivalent to the problem of constructing the roots of the equation
z5−1=0.
One root of this equation is z0=1 which corresponds to the point P0(1,0). Removing the factor corresponding to this root, the other roots turn out to be roots of the equation
z4+z3+z2+z+1=0.
These roots can be represented in the form ω, ω2, ω3, ω4 where ω = exp(2πi/5). Let these correspond to the points P1, P2, P3, P4. Letting
p1=ω+ω4, p2=ω2+ω3
we have
p1+p2 = −1, p1p2=−1. (These can be quickly shown to be true by direct substitution into the quartic above and noting that ω6 = ω, and ω7 = ω2.)
So p1 and p2 are the roots of the quadratic equation
x2+x−1=0.
The Carlyle circle associated with this quadratic has a diameter with endpoints at (0, 1) and (-1, -1) and center at (-1/2, 0). Carlyle circles are used to construct p1 and p2. From the definitions of p1 and p2 it also follows that
p1=2cos(2π/5), p2=2cos(4π/5).
These are then used to construct the points P1, P2, P3, P4.
This detailed procedure involving Carlyle circles for the construction of regular pentagons is given below.[3]
Draw a circle in which to inscribe the pentagon and mark the center pointO.
Draw a horizontal line through the center of the circle. Mark one intersection with the circle as pointB.
Construct a vertical line through the center. Mark one intersection with the circle as point A.
Construct the point M as the midpoint of O and B.
Draw a circle centered at M through the point A. Mark its intersection with the horizontal line (inside the original circle) as the point W and its intersection outside the circle as the point V.
Draw a circle of radius OA and center W. It intersects the original circle at two of the vertices of the pentagon.
Draw a circle of radius OA and center V. It intersects the original circle at two of the vertices of the pentagon.
The fifth vertex is the intersection of the horizontal axis with the original circle.
Regular heptadecagon
There is a similar method involving Carlyle circles to construct regular heptadecagons.[3] The attached figure illustrates the procedure.
Regular 257-gon
To construct a regular 257-gon using Carlyle circles, as many as 24 Carlyle circles are to be constructed. One of these is the circle to solve the quadratic equation x2+x−64=0.[3]
Regular 65537-gon
There is a procedure involving Carlyle circles for the construction of a regular 65537-gon. However there are practical problems for the implementation of the procedure, as, for example, it requires the construction of the Carlyle circle for the solution of the quadratic equation x2+x−214=0.[3]
History
Carlyle's solution to Leslie's problem. The black line segment is divided in two segments in such a way that the two segments form a rectangle (green) being of equal area to another given rectangle (red).
According to Howard Eves (1911–2004) the mathematician John Leslie (1766-1832) described the geometric construction of roots of a quadratic equation with a circle in his book Elements of Geometry and noted that this idea was provided by his former student Thomas Carlyle (1795–1881).[4] However while the description in Leslie's book contains an analogous circle construction, it was presented solely in elementary geometric terms without the notion of a Cartesian coordinate system or a quadratic function and its roots:[5]
To divide a straight line, whether internally or externally, so that the rectangle under its segments shall be equivalent to a given rectangle.
In 1867 the Austrian engineer Eduard Lill published a graphical method to determine the roots of a polynomial (Lill's method). If it is applied on a quadratic function, then it yields the trapezoid figure from Carlyle's solution to Leslie's problem (see graphic) with one of its sides being the diameter of the Carlyle circle. In an article from 1925 G. A. Miller pointed out that a slight modification of Lill's method applied to a normed quadratic function yields a circle that allows the geometric construction of the roots of that function and gave the explicit modern definition of what was later to be called Carlyle circle.[6]
Eves used the circle in the modern sense in one of the exercises of his book Introduction to the History of Mathematics (1953) and pointed out the connection to Leslie and Carlyle.[4] Later publications started to adopt the names Carly circle , Carlyle method or Carlyle algorithm, though in German speaking countries the term Lill circle (Lill-Kreis) is used as well.[7] DeTemple used in 1989 and 1991 Carlyle circles to devise Compass-and-straightedge constructions for regular polygons, in particular the pentagon, the heptacontagon, the 257-gon and the 65537-gon. Ladislav Beran described in 1999, how the Carlyle circle can be used to construct the complex roots of a normed quadratic function.[8]
1 2 See for instance Hornsby, DeTemple or Howard Eves: An Introduction into the History of Mathematics. Holt, Rinehart and Winston, 3rd edition, 1969, p. 73
1 2 John Leslie: Elements of geometry and plane trigonometry: With an appendix, and copious notes and illustrations. Archibald Constable & Co, 3. Ausgabe, 1817, pp. 176, 340 (online copy (Google)). Note that the comment about Carlyle is not contained in earlier editions of the book (1809, 1811). | 677.169 | 1 |
true or false?
Get an answer to your question ✅ "If two lines are perpendicular, then they form four angles that are each 90 degrees. true or false? ..." in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. | 677.169 | 1 |
The Best of Demos with Positive Impact
An Online Resource for Mathematics Instructors
NEW DEMO
Polar Function Gallery
Polar curves are not drawn using the (x, y) coordinates corresponding to horizontal and vertical axes (called Cartesian coordinates).
Instead, polar curves are drawn using polar coordinates. The first of the polar coordinates of a point is r: the distance between the point
and the pole (the pole is another name for the origin (0, 0), and this distance r is often referred to as the radius).
The second of the polar coordinates is \(\mathbf{\theta}\) (read as "theta"): the angle formed by a ray (beginning at the pole extending
to the point) and the polar axis (essentially the x-axis). A polar curve is simply the resulting graph of a polar equation
which is defined by coordinates \(\mathbf{\text{(r,} \theta)}\).
The ordered pairs are graphed on a polar grid as shown in Figure 1.To generate the ordered pairs we use a polar equation \(\mathbf{r = f( \theta)}\)
where r is the value of a polar function evaluated at an angle \(\mathbf{\theta}\).
(A portion of this introduction appeared in Counting Rose Petals.)
A common type of polar curve is a rose curve which has quite a bit of history.
The simplicity and symmetry of Rhodonea or rose curves have fascinated mathematicians since they were first named by the
Italian mathematician Guido Grandi in the 1700s. These curves created fascination due to the interesting patterns for the number of petals of the curves.
The functions in this case are trigonometric expressions \(\mathbf{ \color{green}{r = } \color{red}{a} \color{green}{*cos(} \color{blue}{b} \color{green}{* \theta)} }\)
and \(\mathbf{ \color{green}{r = } \color{red}{a} \color{green}{*sin(} \color{blue}{b} \color{green}{* \theta)}} \).
The number of petals that are present will depend on the value of \(\mathbf{ \color{blue}{b}}\). The value of \(\mathbf{\color{red}{a}}\)
will determine the length of the petals.
There are a wide variety of rose curves. If \(\mathbf{ \color{blue}{b}}\) is an even integer, then the rose will have \(\mathbf{ \color{blue}{2b}}\) petals. If \(\mathbf{ \color{blue}{b}}\)
is an odd integer, then the rose will have \(\mathbf{ \color{blue}{b}}\) petals.
Several rose curves with a various number of petals appears in Figure 2. The animations for rose curves in the gallery below show how things can change using controls.
Polar | 677.169 | 1 |
Elements of Geometry, Geometrical Analysis, and Plane Trigonometry: With an Appendix, Notes and Illustrations
From inside the book
Results 1-5 of 64
Page 31 ... AC is common ; the two tri- angles ABC and ADC are ( I. 2. ) equal ... square or rectangle are all of them right angles ; for the opposite sides ... AC . Because AB is parallel to CD , the alter- nate angles BAC and ACD are ( I. 25 ...
Page 51 ... square described on the hypotenuse of a right - angled triangle , is equivalent to the squares of the two sides . Let ACB be a triangle which is right - angled at B ; the square of the hypotenuse AC is equivalent to the two squares ofPage 53 ... square or rhomboid BF ( II . 1. cor . ) , for it stands on the same base AB and between the same parallels FL and AH ... AC be equivalent to the two squares of AB and BC ; the triangle ABC is right- angled at B. For draw BD perpendicular to ...
Page 55 ... square of the hypotenuse CF is equivalent to the squares of CD and DF ( II . 14 ) , and consequently , taking the ... AC in I. And be- cause the rhomboids KI and AH stand on the same BOOK II . 55 | 677.169 | 1 |
Moment of inertia for right-angle-Ix-Case-1.
What is the difference between case 1 and case 2?
Case no.1 is the one for which the x-axis coincides with the base and the y-axis coincides with the opposite side, while case no.2 is the one for which the x-axis coincides with the base and the y-axis is at the intersection point between the adjacent side and the hypotenuse side.
For Case no.1, the moment of inertia for right-angle-Ix is the second case in the reference attached table from FE Exam reference manual 3-5. While case 2 is the first case in the same table. This is the list of the first moment of area and inertia for the common plane shapes.
The next slide image represents the difference between Case -1 and Case-2 of the right-angle Triangle. For case -1 the opposite of the triangle is at the left side and for case-2, the opposite of the triangle is at the right.
How to get inertia for right-angle-Ix by using a horizontal strip?
For an x-axis passing by the base and The Y-axis passing by the left side, using the following steps, we will estimate the inertia for right-angle-Ix using a horizontal strip: 1-for line CB, we will write the equation as y=-h*x/b+h, we can check the validity of this equation by substituting the value of x coordinates of both C& b and get the corresponding coordinate y value.
2- Our strip is a horizontal strip with width=dy and base=x.
3-dA, which is the area of the horizontal strip will be equal to the base multiplied by the height, x*dy. To get the dy, we will differentiate the y equation, which we have already estimated. 4- Our moment of inertia due to the strip is dIx=dA*y^2, remember that dA=x*dy. 5- We can write dIx=b*(h-y)/h*(y^2*dy), after substitution by the value of dy.
6-the value of Ix=∫dIx=∫b(h-y)/h(y^2dy), from y=0 to y=h.6- after integration and substitution we get Ix=bh^3/12. The value of inertia for a right-angle-Ix=base*height^3/12. the next slide image shows the steps.
How to get the inertia for right-angle-Ix by using a vertical strip?
Using the following steps, we will estimate the Ix the moment of inertia for the right-angle triangle by using a vertical strip:
1- we are going to move this strip horizontally, so we have to substitute the value of y=-h/b*x+h, as we will see later. 2- Our strip is a vertical strip with width=dx and height=y. 3- The strip with an area dA, will be=y*dx.
4-The moment of inertia for the small element -dIx from our study of a rectangular section can be estimated as dx*y^3. 5- The moment of inertia for the small element – dIx value is shown as per the next slide picture.
6-Integrate from x=0 to x=b.
7- We will get the same value of Ix as estimated by using a horizontal strip, which is=bh^3/12.8- k^2x=Ix/A=bh^3/(12(0.50b*h))=h^2/6.
How to get inertia for right-angle-Ix but at the CG?
To get the moment of inertia at the Cg, we can use the expression of Ix and then find Ixg at the CG from the theorem of parallel axes, by deducting the product of area by the square of the distance between Cg and the X-axis.
The moment at the Cg, Ixg=Ix-A*ybar^2. Substitute by the value of Ix, write Ixg=(b*h^3/12)-(0.50*b*h)*(h/3)^2=Ixg=b*h^3/36. The radius of gyration at the Cg, Kxg=sqrt(IxG/Area)=sqrt(bh^3/36)/(0.50bh))=h/(3*sqrt(2)).
This is a link for the next post, Moment of inertia – for the right-angle triangle-case-1-Iy. | 677.169 | 1 |
This lesson will allow students to select and share what details are …
This lesson will allow students to select and share what details are important on a topic. Groups of students will research a topic and then discuss and determine the top 25 important things someone should know about the topicAn interactive applet and associated web page that demonstrate the properties of …
An interactive applet and associated web page that demonstrate the properties of a 30-60-90 triangle. The applet shows a right triangle that can be resized by dragging any vertex. As it is dragged, the remaining vertices change so that the triangle's angles remain 30 degrees, 60 degrees and 90 degrees The text on the page points out that the sides of a 30-60-90 triangle are always in the ratio of 1 : 2 : root 3 Applet can be enlarged to full screen size for use with a classroom projector. This resource is a component of the Math Open Reference Interactive Geometry textbook project at
NOTE: These are video notes! They are to be used AFTER …
NOTE: These are video notes! They are to be used AFTER instruction of these concepts or possibly as a precursor to the actual activities.
The link provided is a 27 minute video that I have used for notes with my 6th grade honors class. Obviously students shouldn't sit down and watch a 27 minute video all at once but it could be used in a part series as students are discovering these concepts in your class with actual activities.
I love Educreations because it's really easy to make a video, get a link that you can embed into any online classroom space, and a large portion of the storage space is free! YAY FREE! task is focused on Common Core Standards 3.OA.3 Students are applying … | 677.169 | 1 |
Practice Introduction to Three dimensional Geometry Class 11 MCQ test. This Introduction to Three dimensional Geometry Class 11 MCQ test check the understanding and concept of the chapter. You can test your knowledge and evaluate yourself. | 677.169 | 1 |
psychiatricethics
A regular octagon rotates 360° about its center. How many times does the image of the octagon coinci...
3 months ago
Q:
A regular octagon rotates 360° about its center. How many times does the image of the octagon coincide with the preimage during the rotation?
Accepted Solution
A:
Answer: 8 timesStep-by-step explanation: By definition, the sum of the exterior angles of a polygon is 360 degrees. Therefore, each exterior angle of the polygon is: [tex]\frac{360\°}{n}[/tex] Where n is the number of sides of the polygon. The lengths of the sides of a regular octagon are equal and the measures of the internal angles are equal too. The sum of the exterior angles of a polygon is 360 degrees and it has 8 sides, therefore, the measure of each exterior angle is: [tex]=\frac{360\°}{8}=45\°[/tex] You know that the regular octagon rotates 360° about its center. Therefore, keeping all the above on mind, you have that the number of times (which you can call x) the image of the octagon coincide with the preimage during the rotation is: [tex]x=\frac{360\°}{45\°}=8\ times[/tex] | 677.169 | 1 |
ryantechinc
Figure 1 is transformed to Figure 3, as shown in the diagram. Describe the transformation. A) dila...
4 months ago
Q:
Figure 1 is transformed to Figure 3, as shown in the diagram. Describe the transformation. A) dilation, then reflection B) reflection, then rotation C) rotation, then reflection D) translation, then rotation | 677.169 | 1 |
A school Euclid, being books i. & ii. of Euclid's Elements, with notes by C ...
The above methods and illustrations will be useful to the student in directing him how to proceed in the investigation of an exercise, but no method can be given which will necessarily succeed in all cases. In such a branch of study only careful observation of the methods employed in different cases and extensive practice in their application can give confidence and
ensure success.
EXERCISES. BOOK I.
Sides and Angles of Triangles.
Conditions of Equality.
Props. 4, 8, 26, 5 and 6.
1. The perpendiculars from the middle points of two sides of a triangle meet in a point which is equidistant from the three angles of the triangle.
2. Find a point within an acute-angled triangle from which a circle may be described about the triangle.
3. Two angles of an equilateral triangle are bisected, and the point at which the bisecting lines intersect is joined with the third angle; show that the third angle will be bisected.
4. The exterior angles at the base of an isosceles triangle ABC are bisected by the straight lines AO, BO intersecting in 0. If 40 be joined show that it will bisect the vertical angle ACB.
5. In any isosceles triangle if a straight line be drawn from the vertex to the base and fulfil any one of the following conditions it must fulfil the other two. The conditions are that the line 1, bisects the base, 2 bisects the vertical angle, 3 is perpendicular to the base. (Three exercises.)
[These properties of an isosceles triangle form the key to the next three exercises.]
6. To find a point in a given straight line which shall be equidistant from two given points.
7. Through a given point to draw a straight line equally inclined to two given lines.
8. From two given points on opposite sides of a line draw two straight lines which shall meet in that given straight line and include an angle bisected by the given straight line.
9. The diagonals of a rhombus bisect each other at right angles.
10. Inscribe a rhombus in a given triangle, having one of its angles coincident with an angle of the triangle.
11. Show that the perpendiculars dropped from any point in the line bisecting an angle of a triangle upon the sides containing that angle, or those sides produced, will be equal to one another.
12. Find a point in a given straight line from which the perpendiculars upon two given straight lines which intersect shall be equal to one another.
Sides and Angles of Triangles.
Conditions of Inequality.
Props. 18, 19, 20, 21, 24, 25.
13. If AD is the longest side and BC the shortest of a quadrilateral ABCD, then the angle ABC is greater than ADC, and the angle BCD greater than BAD.
14. In any quadrilateral figure show that any one side is less than the sum of the other three.
15. The four sides of a quadrilateral are together greater than the sum and less than twice the sum of the diagonals.
16. The two sides of a triangle are together greater than twice the line joining the vertex with the middle point of the base.
17. The sum of the straight lines drawn from any points within a triangle to each of its angles is less than the sum and greater than half the sum of the sides of the triangle.
18. Show that the last exercise is true if a square be substituted for the triangle.
19. Two quadrilateral figures ABCD and AEFD stand on the same base AD and AEFD lies wholly within ABCD. Show that the perimeter of AEFD is less than that of ABCD.
[Produce EF to meet AB and CD in G and H.]
20. The shorter diagonal of a rhombus passes through the greater angle.
21. The perpendicular is the shortest straight line which can be drawn from a point to a given straight line, and of all others one nearer to the perpendicular is less than one more remote, and lines equally remote from the perpendicular are equal to one another.
Parallel Lines. Props. 27 to 31.
22. A line making equal angles with the sides of an isosceles triangle will be parallel to the base.
23. If 40 bisects the base of an isosceles triangle ABC and a straight line be drawn at right angles to it cutting the equal sides AB, AC in P and Q then PQ shall be parallel to the base, and shall also be bisected by AO.
24. If a line bisecting the exterior angle of a triangle be parallel to the base, show that the triangle is isosceles.
25. On a given straight line to describe a triangle equiangular to a given triangle.
26. Through the angular points of an equilateral triangle
straight lines are drawn parallel to the opposite sides. Show that the figure thus formed is an equilateral triangle.
27. From a given point draw a straight line to meet a given straight line and making with it an angle equal to a given angle.
28. From any point O draw lines parallel to the sides of any rectilineal figure, and show that the exterior angles of the figure are together equal to the angles at the point O, i.e., to four right angles.
29. Prove by aid of the last exercise that each of the angles of an equilateral triangle is equal to two-thirds of a right angle. 30. Every parallelogram which has one of its angles a right angle has all its angles right angles.
Value of the Interior and Exterior angles of rectilineal figures.
Prop. 32 and corollaries.
31. If one angle of a triangle be equal to the sum of the other two then it is a right angle; but if it be less than the sum of the other two it is an acute angle; and if greater an obtuse angle.
32. Trisect a right angle.
33. What fraction of a right angle is each of the angles of an isosceles triangle, when the angle at the vertex is four times each of the angles at the base? Shew how to construct such a triangle.
34. Express as a fraction of a right angle the interior angle of each of the following regular figures, viz.: a hexagon, an octagon, and a decagon.
35. Show that three regular hexagons, four squares, or six equilateral triangles will fill up all the space about a point.
36. If the sides of any hexagon be produced to meet the angles formed by their intersection will be together equal to four right angles.
37. The middle point of the hypotenuse of a right-angled triangle is equally distant from the three angles.
38. The straight lines EC, BD bisect the base angles of an isosceles triangle ABC, and meet the opposite sides in D and E. Show that EC is equal to BD and ED parallel to BC.
39. Describe a triangle of given perimeter and having its angles equal to those of a given triangle.
40. The hypotenuse BC of a right-angled isosceles triangle is trisected in D and E; from D and E the straight lines DF and EG are drawn at right angles to BC and meeting the sides in F and G. Show that DEGF is a square.
The Parallelogram. Props. 33 and 34.
41. The diagonals of a rectangle are equal to one another. 42. If the opposite sides of a quadrilateral are equal to one another it is a parallelogram.
43. The diagonals of a parallelogram bisect one another.
44. If the opposite angles of a quadrilateral are equal to one another it is a parallelogram.
45. Through a given point draw a straight line so that the part of it intercepted by two given parallel straight lines may be of a given length.
46. ABCD is a trapezium having the side AB parallel to CD. Through the middle point of BC, PQ is drawn parallel to AD meeting AB and DC (produced if necessary) in P and Q. that the parallelogram AQ is equal to the trapezium ABCD.
Shew
47. Any line passing through the middle point of the diagonal of a parallelogram divides the parallelogram into two equal parts.
48. To bisect a parallelogram by a straight line drawn from a given point in one of its sides.
49. If through the angles of a square straight lines be drawn parallel to its diagonals the figure so formed will be a square having twice the area of the given square.
50. Two of the opposite angles of a parallelogram are joined with the middle points of the opposite sides. Shew that the lines thus drawn will trisect the diagonal of the parallelogram.
Parallelograms and Triangles. Equivalent Areas.
Props. 35 to 45.
51. Bisect a triangle by a straight line drawn from one of its angles.
52. Bisect a triangle (ABC) by a straight line drawn from a given point (P) in one of its sides (AB).
53. The straight line joining the middle points of two sides of a triangle is parallel to the third side and equal to half of it. 54. The straight lines joining the middle points of the four sides of any quadrilateral form a parallelogram.
55. Bisect a square by a straight line drawn from a given point in one of its sides. [See Exercises 47 and 48.]
56. Divide a square into four equal parts by straight lines drawn from a given point in one of its sides.
57. The diagonals of a parallelogram divide it into four equivalent triangles.
58. If lines be drawn from the angles of a parallelogram to any point within the figure the parallelogram will be divided into two pairs of equivalent triangles.
59. Change a quadrilateral figure into an equivalent triangle. 60. If two sides of a triangle are given the area of the triangle is greatest when they contain a right angle.
Relations of the squares described on the sides of a Triangle.
Props. 47 and 48. Book i. Props. 12 and 13. Book ii.
61. Make a square which shall be (1) double of a given square. (2) Equal to the squares described on three given lines. 62. Find a square equal to the difference of two given squares. 63. If a perpendicular be drawn from the vertex of a triangle to the base, the difference of the squares on the segments of the base is equal to the difference of the squares on the sides of the triangle.
64. If a point be taken within or without a rectangle the sum of the squares of the lines drawn to one pair of opposite angles is equal to the sum of the squares on the lines drawn to the other two angles.
65. If two right-angled triangles have the hypotenuse, and one side equal in each triangle, the triangles shall be equal in every respect.
66. An angle of an isosceles triangle is 120° and the sides containing it are each 4 units in length. What is the length of the remaining side?
67. In any right-angled triangle if a perpendicular be drawn from the right angle upon the hypotenuse the square on this line is equal to the rectangle under the segments of the base.
68. In the last exercise if ABC be the triangle and BD the perpendicular, show that the rectangle CA, AD is equal to the square on AB,
69. If the base of a triangle be bisected the squares on the other two sides of the triangle are together double of the square on half the base and of the square on the line, joining the middle point of the base with the opposite angle.
[Draw AD bisecting side BC of the triangle ABC and AE perpendicular to it. Then one of the triangles ADC, ADB is obtuse and the other acute. Suppose ADC obtuse. Then by ii. 12 the square on AC-squares on AD, DC increased by the twice the rectangle CD, DE: and by ii. 13 the square on AB-squares on AD.DB. diminished by twice the rectangle BD, DE (=CD.DE). Hence the sqs. on CA, AB=twice the sqs. on CD and DA.]
70. From the last exercise show that the squares on the sides of a parallelogram are together equal to the squares on its diagonals. | 677.169 | 1 |
Learn Mathematics
Tag: ruleSets – Cartesian Product of Sets If A and B are two non empty sets, then their Cartesian Product A x B is set of all possible ordered pairs. A x B = {(x, y) : x ∈A, y ∈ B} where the elements of A are comes first and the elements of B are […]
Rules of Addition and Subtraction of like Terms In this lesson we will learn, how to add and subtract like terms Like Terms In an algebraic expression, like terms are terms that have same variables with same exponents and only the numerical coefficients of like terms are different. The terms must have the same variables […]
What is Algebra Algebra (from Arabic al-jabr meaning "reunion of broken parts") is the branch of mathematics. The most general form of algebra is the study of mathematical object in abstract algebra is called an algebra. In algebra uses letters like x,y a and b or other symbols in place of values, and the rules […]
Order of mathematical Operations The four basic operation in mathematics, as everyone knows, are addition, subtraction, multiplication and division. These basic operations are used in algebra with numbers or letters or combination of both. A number multiplied by a variable, a number or variable is called a term. An expression is combination of such terms. Example: In […]
Theorem The perpendicular from the center of a circle to a chord bisects the chord. Given: A circle with center O, AB is chord of a circle and OC perpendicular from the center to the chord AB. i.e. OC ⊥ AB therefore ∠OCA and ∠OCB Both angles are 900. To prove: AC = CB (C is the mid point of chord AB) […]
Representation of sets There are three methods of representing a set. 1. Statement form method 2. Roster or tabular form method 3. Rule or Set builder form method Statement form method In this method well defined description of the […] | 677.169 | 1 |
What is a Triangle?
Due to the word "tri," which stands for "three," the polygon with three sides is referred to as a triangle. Additionally, the name of the polygon suggests that it contains three angles. A polygon having three edges & three vertices is called a triangle. It is one of the fundamental geometric shapes.
The symbol for a triangle having vertices A, B, and C is ∆ABC. Triangles can be found in a variety of everyday objects, such as sandwiches, traffic signs, clothes hangers, and billiards racks. The area can be calculated using the formula "a = $\dfrac{bh}{2}$."
Triangle
Area of a Triangle with Three Points
Area of Triangle with 3 Points Formula
In a two-dimensional plane, the area of a triangle is the area that it completely encloses. A triangle is a closed shape with three sides and three vertices, as is common knowledge. The entire area occupied by a triangle's three sides is referred to as its area. Half of the product of the triangle's base and height provides the general formula for calculating the area of the triangle.
Area of a Triangle
Area of Triangle with Coordinates of the Vertices
Triangle Vertices Formula
An ordered pair of real integers, known as the coordinates of that location, can be used to represent a point in a plane (sometimes referred to as a cartesian plane or a coordinate plane). Coordinate geometry is the area of mathematics that deals with employing coordinate systems to solve geometrical problems.
We will look at how to determine the area of the triangle using vertices if the triangle's coordinates are given. The area of the triangle is calculated if the triangle's coordinates are $\left(x_{1}, y_{1}\right), \left(x_{2}, y_{2}\right)$, and $\left(x_{3}, y_{3}\right)$.
Area of a Triangle with Coordinates
Area of $\Delta \mathrm{PQRSample Problems
1. If A(1, 2), B(4, 2), and C(5, 5) are the vertices of the ABC, respectively. What is the area of the ABC?
Ans: Here, the points are given as
$\left(x_{1}, y_{1}\right)=(1,2)$
$\left(x_{2}, y_{2}\right)=(4,2)$
$\left(x_{3}, y_{3}\right)=(5,5)$Substituting the values, we get
$=\dfrac{1}{2}[1(2-5)+4(5-2)+3(2-2)]$
$=\dfrac{1}{2}[(-3)+12+(0)]$
= Area = 4.5
Thus, the area of ABC is 4.5 sq units.
2. Find the area of the triangle formed by the points $(5,2),(-9,-3)$ and $(-3,-5)$
Ans: Let $\mathrm{A}\left(\mathrm{x}_1, \mathrm{y}_1\right)=(5,2), \mathrm{B}\left(\mathrm{x}_2, \mathrm{y}_2\right)=(-9,-3)$ and $\mathrm{C}\left(\mathrm{x}_3, \mathrm{y}_3\right)=(-3,-5)$ be the points of a triangle.3. Find the area of the triangle whose vertices are $(10,-6),(2,5)$ and $(-1,3)$.
Ans: Area of $\triangle A B C=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$
Substituting the values, we get
$A(10,-6), B(2,5)$ and $C(-1,3)$
Area of triangle $=\frac{1}{2}(49)(10(5-3)+2(3+6)-1(-6-5))$
$=\frac{1}{2}(49)(20+18+11)$
$=\frac{1}{2}(49)$
$=24.5 \mathrm{sq}$ units
Thus, the triangle is $=24.5 \mathrm{sq}$ units.
Practice Questions
Let's solve the following area of the triangle with the formula.
1. What is the area of the triangle whose vertices are X(3, 3), Y(3, 4), and Z(7, 0)?
Ans: 2 sq units
2. What is the area of the triangle whose vertices are A(4, 1), B(4, 5), and C(5, 6)?
Ans: 2 sq units
3. What is the area of the triangle whose vertices are A(1, 3), B(2, 2), and C(3, 5)?
Ans: 2 sq units
4. What is the area of the triangle whose vertices are X(1, 6), Y(4, 2), and Z(8, 5)?
Ans: 12.5 sq units
5. What is the area of the triangle whose vertices are A(4, 4), B(2, 2), and C(5, 5)?
Ans: 0 sq units
Summary
Generally speaking, an "area" is the area included within the perimeter of a flat object or figure. Measurements are made in square units, with square metres serving as the reference unit (m2). There are predefined formulas for the computation of area for squares, rectangles, circles, triangles, etc.
This article taught us the formulas for the area of triangles for various kinds of triangles, along with several sample issues. As triangles are isosceles and equilateral, they all have the same formula. The formula for the area of a triangle can be determined by using the Pythagorean theorem.
FAQs on Area of Triangle With 3 Points
The different types of triangles include an equilateral triangle, an isosceles triangle as well as a right-angled triangle.
2. What are the different properties of the triangle?
Three sides and three angles make up a triangle. A triangle's total angles are always 180 degrees. A triangle's outside angles are always a sum of 360 degrees. Successive interior and exterior angles added together are supplementary. Any two sides of a triangle can be added together to have a length larger than the third side.
3. What are the angles of the triangle?
In a triangle, there are three angles. Two of the triangle's sides are known as the vertex that meets at a common location to form these angles. Three inner angles added together equal 180 degrees. The side length generates an outside angle if we stretch it outside. | 677.169 | 1 |
Geometry: I want to understand some points
In summary, the conversation discusses finding the unknown value (x) when given the length of an arc and other variables such as the radius (R) and angle (alpha). The figure attached shows a triangle and the formula R = xR + (1-x)R is suggested to find the unknown value. The term "recedes from its chord" refers to the arc moving away from the chord. The friend suggests using x = R - R cos (alpha) to find the value and explains that it represents the greatest distance between the arc and chord. The conversation also mentions using the properties of a triangle to find the angle and radius.
Jul 28, 2005
#1
irony of truth
90
0
What is my unknown (say x) if I am finding the value in which an arc of a great circle 10 units long recedes from its chord? (given radius R and angle alpha?)
He has given the greatest distance between the arc and the chord i.e. the difference of the radius and distance of the chord from the center
Jul 30, 2005
#6
gopi krishna
1
0
hi
By using the laws properties of triangle you will get the angle to be 45 degrees and
R to be square root of 50
Last edited by a moderator: Apr 21, 2017
What is geometry?
Geometry is a branch of mathematics that focuses on the study of shapes, sizes, and positions of objects in space.
What are the basic elements of geometry?
The basic elements of geometry include points, lines, and angles. Points are represented by a dot and have no size or dimension. Lines are straight and extend infinitely in both directions. Angles are formed by two intersecting lines or line segments.
What is the difference between 2D and 3D geometry?
2D geometry deals with shapes and figures that are flat and have only two dimensions - length and width. 3D geometry, on the other hand, involves objects that have three dimensions - length, width, and depth.
What are some applications of geometry in real life?
Geometry has many practical applications in our daily lives, such as in architecture, engineering, navigation, art, and design. It is also used in fields like astronomy, physics, and computer graphics.
How can I improve my understanding of geometry?
To understand geometry better, you can start by studying the basic concepts and principles, practicing with different problems, and visualizing shapes and figures in your mind. You can also use online resources, textbooks, or seek help from a tutor or teacher. | 677.169 | 1 |
The first six books of the Elements of Euclid, with numerous exercises
Im Buch
Ergebnisse 1-5 von 20
Seite 2 ... perpendicular to it . XI . An obtuse angle is that which is greater than a right angle . XII . An acute angle is that which is less than a right angle . Obtuse . Acute . XIII . A term or boundary is the extremity of any thing . XIV . A ...
Seite 11 ... perpendicular to a b from the point c . Take any point d upon the other side of a b , and from the centre c , at the distance cd , describe ( 3 post . ) the circle egf meet- ing a b in f , gi and bisect ( i . 10 ) fg in h , and join cf ...
Seite 34 ... perpendicular to a given line is the shortest that can be drawn from the point to the line . 17. If a triangle have its exterior angle and one of its opposite interior angles double of those in another triangle , its remaining opposite ...
Seite 43 ... perpendicular falls , and the straight line in- tercepted without the triangle between the perpendicular and the obtuse angle . LET abc be an obtuse - angled triangle , having the obtuse angle acb , and from the point a let ad be drawn ...
Seite 44 ... perpendicular ( i . 12 ) ad from the opposite angle : the square of a c , opposite to the angle b , is less than the ... perpendicular to bc ; then is bc the straight line between the perpendicular and the acute angle at b ; and it is ...
Beliebte Passagen
Seite 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another : XVI.
Seite 21Seite 15 - If, from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than, the other two sides of the triangle, but shall contain a greater angle. Let... | 677.169 | 1 |
Question 4 - Area of Parallelograms - Areas of Parallelograms and Triangles
Last updated at April 16, 2024 by Teachoo
Transcript
Question 4
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
ar (APB) + ar (PCD) = 1/2 ar (ABCD)
Given: A parallelogram ABCD
To prove: ar (APB) + ar(PCD) = 1/2 ar (ABCD)
Proof: Since ABCD is a parallelogram
AB ∥ CD & AD ∥ BC
We draw line EF passing through P, parallel to AB & DC
i.e. EF ∥ AB ∥ CD
Here, AE ∥ BF
& AB ∥ EF
∴ EFBA is a parallelogram
Similarly, EFCD is a parallelogram
Now,
ΔAPB and parallelogram EFBA
are on the same base AB
and between the same parallel lines AB and EF,
∴ Area (ΔAPB) = 1/2 Area (EFBA)
Similarly,
ΔPCD and parallelogram EFCD
are on the same base CD
and between the same parallel lines CD and EF,
∴ Area (ΔPCD) = 1/2 Area (EFCD)
Adding (1) & (2), we obtain
Area (∆APB) + Area (∆PCD) = 1/2 Area (EFAB) + 1/2 Area (EFCD)
Area (ΔAPB) + Area (ΔPCD) = 1/2 Area (ABCD)
Question 4
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
Given: A parallelogram ABCD
To prove: ar (APD) + ar(PBC) =ar (APB) + ar (PCD)
Proof: Since ABCD is a parallelogram
AB ∥ CD & AD ∥ BC
We draw line GH passing through P, parallel to AD & BC
i.e. GH ∥ AD ∥ BC
Here, GH ∥ AD
& AG ∥ DH
∴ AGHD is a parallelogram
Similarly, GBCH is a parallelogram
Now,
ΔAPD and parallelogram AGHD
are on the same base AD
and between the same parallel lines AD and GH,
∴ Area (ΔAPD) = 1/2 Area (AGHD)
Similarly,
ΔPBC and parallelogram GBCH
are on the same base BC
and between the same parallel lines BC and GH,
∴ Area (ΔPCB) = 1/2 Area (GBCH)
Adding (1) & (2), we obtain
Area (∆APD) + Area (∆PBC) = 1/2 Area (AGHD) + 1/2 Area (GBCH)
Area (∆APD) + Area (∆PBC) = 1/2 Area (ABCD)
Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)
Hence proved | 677.169 | 1 |
Double angle identity - Tr
What is a double angle? Double angle, as the words imply, means to increase the size of the angle to twice its size. It simply means two times of a trigonometric angle i.e. 2x in terms of x. Double-angle identities are used to simplify trigonometric calculations. Double angle formula. The main formulas used to find a double angle are: You may know this stigmatized condition as multiple personality disorder or split personality. It's real and treatable. Here are the main DID signs and symptoms. Dissociative ident...Feb 19, 2018 · This is the first double angle formula for cosine. To get another formula, we first need to reflect on a Pythagorean Identity . We can manipulate it by subtracting sin 2 x from both sides to get... If we take this expression for cos 2 x and replace it within our first double angle formula for cosine, this is the result. LetJun 1, 2022The half-angle formulas tell you how to find the sine or cosine of x/2 in terms of the sines and cosines of x. They follow from the double-angle formulas.Jan Using Cisgender, transgender, nonbinary, no gender, and others — we look at some of the many identity terms people may use to describe their gender. Gender identity is your personal expe...We separated the fractional term because we notice we have a double angle. Recalling our trigonometric identities, the fractional term is the inverse of the power reducing formula for sine. Now separating out the sine terms: Now recalling the basic identities: Using the trigonometric identities we have proven that the equation is true. Double Angle Trigonometric Identities. If the angles are doubled, then the trigonometric identities for sin, cos and tan are: sin 2θ = 2 sinθ cosθ; cos 2θ = cos 2 θ – sin 2 θ = 2 …If you're a brand marketer, designer, developer, or otherwise, you need a visual identity system for your organization. Here's what it is and how to make one. Trusted by business b...Example 4: Using Double-Angle Identities to Solve a Trigonometric Equation. Find the solution set for 𝑥 given c o s c o s 2 𝑥 + 1 3 √ 3 𝑥 = − 1 9, where 𝑥 ∈] 0, 2 𝜋 [. Answer . In this example, we are going to solve a trigonometric equation in a particular range using the double-angle identities.In Then the tangent identity just follow from those two and the quotient identity for tangent.Learn how to work with the Double Angle Formulas for sine, cosine, and tangent in this free math video tutorial by Mario's Math Tutoring.0:11 Double angle fo... InUsing the half‐angle identity for the cosine, Example 3: Use the double‐angle identity to find the exact value for cos 2 x given that sin x = . Because sin x is positive, angle x must …Identity theft can be a serious problem, potentially impacting your credit report for years to come. By detecting it early and putting measures in place to prevent it, you can redu...Feb 16, 2023 · Double Cos Double angle identity. The cosine of double angle is equal to the subtraction of square of sine from square of cosine of angle. ( 1). cos 2 θ = cos 2 θ − sin 2 θ. ( 2). cos 2 x = cos 2 x − sin 2 x. Learn more.The doubleangle identities and halfangle identities are special examples of the sum and difference formulae for sine and cosine, respectively. sin2a=2sinacosa. If you want to get all of the basic double angle identities in one place and calculate them quickly, try our double angle formula calculator. Because you'll need to know and remember ...This way, if we are given θ and are asked to find sin (2 θ), we can use our new double angle identity to help simplify the problem. Let's start with the derivation of the double angle identities. One of the formulas for calculating the sum of two angles is: Jul 4, 2023 · Apr 13, 2022 · The double angle formulas are used to find the values of double angles of trigonometric functions using their single angle values. Also, the double-angle formulas can be used to derive the triple-angle formulas. Related Topics. Pythagorean Identities; A step-by-step guide to double angle formulas. The double angle formulas are the special casesWorked example 7: Double angle identities. If \(\alpha\) is an acute angle and \(\sin \alpha = \text{0,6}\), determine the value of \(\sin 2 \alpha\) without using a calculator. Draw a sketch. We convert \(\text{0,6}\) to a fraction …The double angle identities. sin(2α) = 2sin(α)cos(α) cos(2α) = cos2(α) − sin2(α) = 1 − 2sin2(α) = 2cos2(α) − 1. These identities follow from the sum of angles identities. …LetSubA double-angle function is written, for example, as sin 2θ, cos 2α, or tan 2x, where 2θ, 2α, and 2x are the angle measures and the assumption is that you mean sin(2θ), cos(2α), or tan(2x).Because tangent is equal to the ratio of sine and cosine, its identity comes from their double-angle identities. The double-angle identities find the function for twice …The cotangent of a double angle is equal to the reciprocal of the tangent of the double angle. To calculate the cotangent of a double angle, you can use the following formula: cot (2θ) = 1 – tan² (θ) 2tan (θ) The cotangent of 2 times an angle θ is equal to 1 minus the tangent squared of the angle θ divided by 2 times the tangent of the ...Solve Trigonometric Equations. We can use the half and double angle formulas to solve trigonometric equations. Let's solve the following trigonometric equations. x = 0 when 0 ≤ x < 2π 0 ≤ x < 2 π. Change tan 2x tan 2 x and simplify. tan 2x + tan x 2 tan x 1 −tan2 x + tan x 2 tan x + tan x(1 −tan2 x) 2 tan x + tan x −tan3 x 3 tan x ...If you're a brand marketer, designer, developer, or otherwise, you need a visual identity system for your organization. Here's what it is and how to make one. Trusted by business b...If you want to view all of my videos in a nicely organized way, please visit . It is free for anyone to use. Please share with...2Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side. Example 3. Using the Double-Angle Formulas to Establish an Identity.The double angle identities are trigonometric identities that give the cosine and sine of a double angle in terms of the cosine and sine of a single angle. sin(2θ) = 2sin(θ)cos(θ) cos(2θ) = cos2(θ)−sin2(θ) Two additional double angle identities are sometimes given that combine the pythagorean identity and the cosine identity above using ...Identive is presenting Q4 earnings on March 2.Wall Street predict expect Identive will report losses per share of $0.004Follow Identive stock pric... On March 2, Identive will be r...cos(2θ) = cos2 θ −sin2 θ = 2cos2 θ − 1 = 1 − 2sin2 θ. The double-angle identity for the sine function uses what is known as the cofunction identity. Remember …A more valuable company than Apple or Amazon—for now. Microsoft has a real shot to end the year as the most valuable public company in the world. That wasn't the case a year ago, aCisgender, transgender, nonbinary, no gender, and others — we look at some of the many identity terms people may use to describe their gender. Gender identity is your personal expe...An identity crisis is an opportunity to grow. Here are the main symptoms and how to cope to overcome identity confusion. There are a few definitions for identity crisis, but the co...Learn how to use double angle identities to solve equations, find the exact values of expressions, and simplify expressions. See the double angle identities table, examples, and questions with answers on this web page. Mar The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself. From these formulas, we also have the following identities:Learn about social identity from this article. Visit HowStuffWorks.com to learn more about what social identity is. Advertisement Social identity relates to how we identify ourselv...Voiceover: In the last video we proved the angle addition formula for sine. You could imagine in this video I would like to prove the angle addition for cosine, or in particular, that the cosine of X plus Y, of X plus Y, is equal to the cosine of X. Cosine of X, cosine of Y, cosine of Y minus, so if we have a plus here we're going to have a ...Doublet. e. In trigonometry, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables for which both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are To show you where the first of the double-angle identities for cosine comes from, this example uses the angle-sum identity for cosine. Because the two angles are equal, you can replace β with α, so cos (α + β) = cosα cosβ – sin sinβ becomes. To get the second version, use the first Pythagorean identity, sin 2 + cos 2 = 1How do you find the value of #tan60# using the double angle identity? Trigonometry Trigonometric Identities and Equations Double Angle Identities. 1 Answer Nghi N Oct 18, 2017 #sqrt3# Explanation: Use trig identity: #tan 2a = (2tan a)/(1 - tan^2 a)# In this case: # ...By the Pythagorean Identity, $\,\cos^2 x = 1 - \sin^2 x\,$ and $\,\sin^2 x = 1 - \cos^2 x\,.$ Thus, we get two alternative versions of the cosine double angle formula ...Author: Roger. Revision notes on 5.6.2 Double Angle Formulae for the Edexcel A Level Maths: Pure syllabus, written by the Maths experts at Save My Exams.This video shows you the basics of Double Angle Trig Formulas. It will show you where they come from and how to use them. cos 2A = cos 2 A − sin 2 A, cos 2A = 2cos 2 A − 1, cos 2A = 1 − 2sin 2 A This video shows you how to use double angle formulas to prove identities as well as derive and use the double angle tangent identity.The double angle identities examples with solutions to learn how to expand the trigonometric functions having double angles and also learn how to simplify the expressions as the trigonometric functions with double angles. Evaluate 1 sin 10 ∘ − 3 cos 10 ∘. Learn solution. Evaluate − sin x + sin 2 x 1 − cos x + cos 2 x. Learn solution.Simplifying trigonometric functions with twice a given angle. Add to Library. Details. Resources. Download. Quick Tips. Notes/Highlights.TrDouble Angle Trigonometry identity calculator is an online tool for computing problems related to trigonometry double angle identities. This calculator can be used for all double angle identities like sin 2 theta, cos 2 theta, tan 2 ...Identive is presenting Q4 earnings on March 2.Wall Street predict expect Identive will report losses per share of $0.004Follow Identive stock pric... On March 2, Identive will be r...Double\] The double-angle identity for the sine function uses what is known as the cofunction identity. Remember that, in a right triangle, the sine of one angle is the same as the cosine of its complement (which is the other acute angle). This is because the adjacent side for one angle is the opposite side for the other angle. Subsec 2A. Show Step-by-step Solutions. Derive and use Half-Angle Identities. The derivations of the half-angle identities for both sine and cosine, plus listing the tangent ones. Then a couple of examples using the identities. Examples: Find the exact value for sin (9π/8) Find cos x/2 if sin x = -4/5 with 3π/2 < x < 2π. Show Step-by-step ...Compound angle formula can be extended to find related double, triple of nth angle formula. Anil Kumar Math Classes:...Double angle calculator used to calculate Double angle formula which refers to the expression of trigonometric functions of angles equal to 2θ in terms of θ. The double angle identity formula is an equation that expresses a trigonometric function of twice an angle in terms of trigonometric functions of the angle.An identity crisis is an opportunity to grow. Here are the main symptoms and how to cope to overcome identity confusion. There are a few definitions for identity crisis, but the co...Identity theft is the fastest growing crime in the U.S. Learn about Internet identity theft, credit card fraud and identity theft protection. Advertisement You work hard every day ...2 days ago · Subject classifications. Formulas expressing trigonometric functions of an angle 2x in terms of functions of an angle x, sin (2x) = 2sinxcosx (1) cos (2x) = cos^2x-sin^2x (2) = 2cos^2x-1 (3) = 1-2sin^2x (4) tan (2x) = (2tanx)/ (1-tan^2x). (5) The corresponding hyperbolic function double-angle formulas are sinh (2x) = 2sinhxcoshx (6) cosh (2x ... This trigonometry video tutorial provides a basic introduction on verifying trigonometric identities with double angle formulas and sum & difference identiti... The double angle identity of Cos2x further expands on the basic identity. It provides us with three equivalent forms: cos(2x) = cos²x – sin²x = 2cos²x – 1 = 1 – 2sin²x. These identities illustrate the remarkable versatility of Cos2x, allowing us to express it in terms of a single trigonometric function—either cosine or sine. ...Double Angle Formulae. sin(A + B) = sinAcosB + cosAsinB Replacing B by A in the above formula becomes: sin(2A) = sinAcosA + cosAsinA. so: sin2A = 2sinAcosA. similarly: cos2A = cos 2 A - sin 2 A. Replacing cos 2 A by 1 - sin 2 A in the above formula gives: cos2A = 1 - 2sin 2 A. Replacing sin 2 A by 1 - cos 2 A gives: cos2A = 2cos 2 A - 1. It can ...The Trigonometric Double Angle identities or Trig Double identities actually deals with the double angle of the trigonometric functions. For instance, Sin2 (α) Cos2 (α) Tan2 (α) Cosine2 (α) Sec2 (α) Cot2 (α) Double Angle identities are a special case of trig identities where the double angle is obtained by adding 2 different angles
Cos2x is an important identity in trigonometry which can be expressed in different ways. It can be expressed in terms of different trigonometric functions such as sine, cosine, and tangent.Cos2x is one of the double angle trigonometric identities as the angle in consideration is a multiple of 2, that is, the double of x.According to our study, residents of states like Florida, Delaware, Maryland and New Jersey may be at a higher risk for identity theft. Residents of these .. Calculators Helpful Gu...Does a smartphone raise your risk of identity theft? Learn why and how to protect yourself from HowStuffWorks. Advertisement Here's a scary question: What would happen if someone s...cIdentive is presenting Q4 earnings on March 2.Wall Street predict expect Identive will report losses per share of $0.004Follow Identive stock pric... On March 2, Identive will be r...482 Chapter 7 The left side of this equation almost looks like the result of the double angle identity for sine: sin( )2 sin T Tcos Multiplying both sides of our equation by 2, 1002 2(900)(4.9) 2cos(T)sin(T) Using the double angleTheCos 75° = Sin 15°. So : 2Cos 75° . Sin 75° = 2 Sin 15°Cos°15 = Sin 30°. Upvote • 0 Downvote. Add comment. Report . | 677.169 | 1 |
An Easy Introduction to the Mathematics: In which the Theory and Practice are Laid Down and Familiarly Explained ... A Complete and Easy System of Elementary Instruction in the Leading Branches of the Mathematics; ... Adapted to the Use of Schools, Junior Students at the Universities, and Private Learners, Especially Those who Study Without a Tutor. In Two Volumes, Volumen2
Dentro del libro
PÃgina 214 ... angles . Pythagoras " was another eminent Grecian philosopher , who ing his ... two treatises on the Tropics and Equi- noxes ; and Suidas , a work on Meteors ... with an equal thirst for useful knowledge , 214 PART VIII . INTRODUCTION .
PÃgina 244 ... with three legs , for taking three points at once . 6. Turn - up Compasses are the plain compasses , with two additional points fixed near the bottom of the legs , the one carrying a port crayon , and the other a drawing pen ; these are ...
PÃgina 245 ... one side , a line of 6 inches , a line of 50 equal parts , and a diagonal scale . On the other side it has a line of chords marked C , and seven decimal scales of various sizes . The line or scale of inches has each inch divided into 10 ...
PÃgina 252 ... angles ; but they agree in one general character , they are both figures . A lion and a sheep differ widely from each other ... have no existing pattern or architype : we can form in the mind the abstract ... two sides of a triangle are ...
PÃgina 264 ... have been omitted . In the ninth , where " a plane rectilineal angle " is ... one and the same plane , as is proved in the second proposition of the ... two angles are formed at the point where they meet ; if these angles be equal to each other ...
PÃgina 281 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
PÃgina 294 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. | 677.169 | 1 |
Erstaunlich Geometry Regents Answer Key 2023 Ideen
Geometry 2010 June Regents Short Response Answers YouTube from
Geometry Regents Answer Key 2023: Your Ultimate Guide Geometry Regents is a standardized test administered in the state of New York to assess students\' knowledge of geometry. The exam is usually taken by high school students in their junior or senior year. The Geometry Regents exam is an essential component of New York State\'s high school graduation requirements. In this article, we will discuss the Geometry Regents Answer Key 2023. The Geometry Regents exam consists of multiple-choice questions, constructed-response questions, and an extended constructed-response question. The exam covers various topics such as circles, triangles, quadrilaterals, coordinate geometry, transformations, and more. It is essential to know the answer key to the Geometry Regents exam to assess your performance and improve your score.
What is the Geometry Regents Answer Key?
The Geometry Regents Answer Key is a document that contains the correct answers to all the questions on the Geometry Regents exam. The answer key is released by the New York State Education Department after the exam is administered. The answer key is an essential resource for students, teachers, and parents to assess the students\' performance on the exam.
When will the Geometry Regents Answer Key 2023 be released?
The Geometry Regents Answer Key 2023 will be released by the New York State Education Department a few days after the exam is administered. The exact date of the answer key release is usually not announced, but it is typically within a week after the exam.
How can you access the Geometry Regents Answer Key 2023?
The Geometry Regents Answer Key 2023 will be available on the official website of the New York State Education Department. The answer key will be in a PDF format that can be downloaded and printed. Students, teachers, and parents can access the answer key to assess the students\' performance on the exam.
Why is the Geometry Regents Answer Key 2023 important?
The Geometry Regents Answer Key 2023 is essential for students, teachers, and parents to assess the students\' performance on the exam. The answer key helps students to identify their mistakes and improve their performance. Teachers can use the answer key to assess the students\' understanding of the concepts and identify areas that need improvement. Parents can use the answer key to monitor their child\'s progress and provide additional support if needed. In conclusion, the Geometry Regents Answer Key 2023 is an essential resource for anyone who has taken the Geometry Regents exam. The answer key helps students, teachers, and parents to assess the students\' performance, identify mistakes, and improve their understanding of the concepts. It is essential to access the answer key to improve your score and achieve success on the Geometry Regents exam. | 677.169 | 1 |
The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :
A
$${x^2} + 12{y^2} = 16$$
B
$$4{x^2} + 48{y^2} = 48$$
C
$$4{x^2} + 64{y^2} = 48$$
D
$${x^2} + 16{y^2} = 16$$
2
AIEEE 2008
MCQ (Single Correct Answer)
+4
-1
A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is :
A
$${{8 \over 3}}$$
B
$${{2 \over 3}}$$
C
$${{4 \over 3}}$$
D
$${{5 \over 3}}$$
3
AIEEE 2006
MCQ (Single Correct Answer)
+4
-1
In the ellipse, the distance between its foci is $$6$$ and minor axis is $$8$$. Then its eccentricity is :
A
$${3 \over 5}$$
B
$${1 \over 2}$$
C
$${4 \over 5}$$
D
$${1 \over {\sqrt 5 }}$$
4
AIEEE 2005
MCQ (Single Correct Answer)
+4
-1
An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F$$' its focii and theangle $$FBF$$' is a right angle. Then the eccentricity of the ellipse is : | 677.169 | 1 |
Question 5. (i) If (bar{a}) and (bar{b}) are any two vectors, then axb is (1) (a) a vector on the same plane where (bar{a}) and (bar{b}) lie. (b) ab cosθ, if θ is the angle between them. (c) a vector parallel to both (bar{a}) and (bar{b}). (d) a vector perpendicular to both (bar{a}) and (bar{b}). (ii) Let (bar{a}) = 2i + 4j – 5k, (bar{b}) = i + 2j + 3k. Then find a unit vector perpendicular to both (bar{a}) and (bar{b}). (2) (iii) Find a vector of magnitude 5 in the direction perpendicular to both (bar{a}) and (bar{b}) (1) Answer: (i) (d) a vector perpendicular to both (bar{a}) and (bar{b}).
2. From the above explanation we have, and parallel. Similarly, |(overline{S P})| = |(overline{R Q})| Therefore, PQRS is a parallelogram.
3. Let (bar{a}) = a1 i + a2 j + a3 k
We hope the given Plus Two Maths Chapter Wise Questions and Answers Chapter 10 Vector Algebra will help you. If you have any query regarding Plus Two Maths Chapter Wise Questions and Answers Chapter 10 Vector Algebra, drop a comment below and we will get back to you at the earliest. | 677.169 | 1 |
21
Página 17 ... ; join DE , and upon ...
Página 24 ... given straight lines , A , B C. SCHOLIUM . If one of the sides were greater than the sum of the other two , the arcs ... rectilineal angle equal to a given rectilineal angle . Let AB be the given straight line , and A the given point in ...
Página 30 ... given point A parallel to the given straight line BC . PROP . XXXII . THEOR . If a side of any triangle be produced ... rectilineal figure are equal to twice as many right angles as the figure has sides , wanting four right angles . For any ...
Página 31 ... rectilineal figure are together equal to four right angles . Because every interior angle ABC , with its adjacent ... given , or merely their sum , the third will be found by subtracting that sum from two right angles . COR . 4. If two ...
Página 37 ... given rectilineal angle . Let ABC be the given triangle , and D the given rectilineal angle . It is required to describe a parallelogram that shall be equal to the given tri- angle ABC , and have one of its angles equal to D. F G Bisect | 677.169 | 1 |
Property of circle regarding chaord and tangent
On the screen, we see a circle with center A and T is a point on its circumference. A tangent is drawn at point T and there is a point P outside the circle, lying on this tangent. From this point P a line is drawn (as shown in the figure) which cuts the circle at points C and D. Let us move the points A, D, P and T and observe how the lengths of PC, DP and PT get changed. | 677.169 | 1 |
Multiple angles worksheet pdf
Printable geometry worksheets for measuring angles with a protractor and drawing. Angles multiple rays sheet 1 math worksheets 4 kids. Use this pdf worksheet to get a picture of how an acute, right, and obtuse angle looks like. Create the worksheets you need with infinite geometry. Most worksheets require students to identify or analyze acute, obtuse, and right angles. The more advanced worksheets include straight and reflex angles too. Turn your pdf or hard copy worksheet into an editable digital worksheet.
You can select different variables to customize these angles worksheets for your. These angles worksheets are perfect for learning and practicing various types problems about different angle topics. Angles multiple rays find the measure of given angles using protractor. Classify everyday objects based on the indicated angle as acute, obtuse, and right angles with this collection of activity worksheet pdfs. Geometry worksheets angles worksheets for practice and study. Classifying angles worksheets math worksheets 4 kids. Sorting angles in multiple rays work your way through these printable sorting angles in multiple rays worksheets by identifying the angles formed. Level 2 has all angles in level 1 in addition to reflex angle. Each worksheet comprises two problems each of which consists of four questions to test your understanding. Each worksheet has 12 problems classifying types of angles. Equations and multipleangle identities kuta software. Level 1 has acute, right, obtuse and straight angles. Multipleangle identities date period kuta software llc.
Study the given images carefully and classify each angle according to. This page has a set of printable worksheets on finding multiples of numbers. Angle measurement worksheets super teacher worksheets. Complementary and supplementary pairs adjacent and nonadjacent angles multiple rays grab these pdf worksheets to demonstrate greater skills in finding the complementary and supplementary angles. Students get to practice measuring angles using builtin protractors in this fun geometry activity.
Kindergarten, 1st grade, 2nd grade, 3rd grade, 4th. Equip children with our classifying angles worksheets, and enable them to identify. Each worksheet has 11 problems identifying the quantity of acute, obtuse right or reflex angles in a shape. This page has printable geometry pdfs on angle types. We also have least common multiple lcm and factoring worksheets. | 677.169 | 1 |
Declaration
Declaration
Determines if the given coordinate falls within any of the polygons.
The optional parameter ignoreBoundary will result in the method returning true if the given coordinate
lies on the boundary line of the polygon or its interior rings. | 677.169 | 1 |
How to Find Arc Length
An arc is any portion of the circumference of a circle. Arc length is the distance from one point to another. Arc length is the distance from one endpoint to the arc of the other. It requires knowing a bit about the geometry of a circle. Here is a step by step guide on how to find arc length.
How to find arc length using a measurement of Central angle in Degrees
Set up the formula for arc length. Arc length = 2π(r)(Θ/360) where r equals the radius of the circle and Θ equals the measurement of the arc's central angle, in degrees.
Plug the given length of the circle's radius into the formula. For example, if the circle's radius is 20 cm, your formula will look like this: Arc length = 2π(20)(Θ/360)
Insert the given value of the arc's central angle into the formula. For example, if the arc's central angle is 145 degrees, your formula will look like this: Arc length = 2π(20)(145/360)
Multiply the radius by 2π. For example 2π(20)(145/360) 2(3.14)(20)(145/360) (125.6)(145/360) | 677.169 | 1 |
Read RD Sharma Solutions Class 9 Chapter 15 Area of Parallelograms and Triangles 15.1
Question 1: Which of the following figures lie on the same base and between the same parallel. In such a case, write the common base and two parallels:
Solution:
(i) ∆APB and trapezium ABCD are on the same base AB and between the same parallels AB and DC.
Here, Common base is AB and Parallel lines are AB and DC.
(ii) Parallelograms ABCD and APQD are on the common base AD and between the same parallels AD and BQ. Here, Common base is AD and Parallel lines are AD and BQ.
(iii) The base of parallelogram ABCD and ΔPQR is different but lies between the same parallels AD and BC.
(iv) ΔQRT and parallelogram PQRS are lies between same parallels QR and PS and on the same base QR.
Here, Common base is QR and Parallel lines are QR and PS.
(v) The base of parallelograms PQRS and trapezium SMNR are on common base SR, but not on same parallels.
It is given that, diagonals AC and BD of a quadrilateral ABCD intersect each other at P.
BQ and DR are two perpendiculars on AC.
ar(ΔAPB) × ar(ΔCDP) = ar(ΔAPD) × ar(ΔBPC)
(1/2×AP×BQ) × (1/2×PC×DR) = ar(ΔAPD) × ar(ΔBPC)
(1/2×PC×BQ) × (1/2×AP×DR) = ar(ΔAPD) × ar(ΔBPC)
ar(ΔAPD) x ar(ΔBPC) = ar(ΔAPD) × ar(ΔBPC)
L.H.S. = R.H.S.
Hence proved
Question 10: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).
Solution:
Draw two perpendiculars CP and DQ on AB.
ar(ΔABC) = 1/2 × AB × CP__________(1)
ar(ΔABD) = 1/2×AB×DQ __________ (2)
In triangles, ΔCPO and ΔDQO
∠CPO = ∠DQO (each 90°)
CO = OD (Given)
∠COP = ∠DOQ (Vertically opposite angles)
ΔCP0 ≅ ΔDQO (By AAS condition)
CP = DQ _________(3) (CPCT)
From equations (1), (2) and (3)
ar(ΔABC) = ar(ΔABD)
Hence proved
Question 11: If P is any point in the interior of a parallelogram ABCD, then prove that area of the triangle APB is less than half the area of parallelogram.
Solution:
Construct DN ⊥ AB and PM ⊥ AB
ar(∥gm ABCD) = AB × DN,
ar(ΔAPB) = 1/2(AB × PM)
PM < DN
AB × PM < AB × DN
1/2 (AB × PM) < 1/2 (AB × DN)
ar(ΔAPB) < 1/2 ar (∥gm ABCD)
Hence Proved
Question 12: If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the mid-point of the median AD, prove that ar(ΔBGC) = 2ar(ΔAGC).
Solution:
It is given that, AD is the median of ΔABC
Construct AM ⊥ BC
BD = DC__________(1)
Multiply AM on both sides of equation (1)
BD × AM = DC × AM
1/2 (BD × AM) = 1/2 (DC × AM)
ar(ΔABD) = ar(ΔACD) _________(2)
In ΔBGC, GD is the median
ar(ΔBGD) = ar(ΔCGD) _________ (3)
In ΔACD, CG is the median
ar(ΔAGC) = ar(ΔCGD) _________ (4)
From (3) and (4) we get,
ar(ΔBGD) = ar(ΔAGC)
ar(ΔBGC) = 2ar(ΔBGD)
ar(ΔBGC) = 2ar(ΔAGC)
Hence Proved
Question 13: A point D is taken on the side BC of a ΔABC, such that BD = 2DC. Prove that ar(ΔABD) = 2ar(ΔADC).
Solution:
It is given that, In ΔABC, BD = 2DC
Take a point E on BD such that BE = ED
BE = ED = DC
We know that median of triangle divides it into two equal triangles.
In ΔABD,
AE is the median.
ar(ΔABD) = 2ar(ΔAED)_________(1)
In ΔAEC,
AD is the median
ar(ΔAED) = 2ar(ΔADC) _________ (2)
From equation (1) and (2) we get,
ar(ΔABD) = 2ar(ΔADC)
Hence proved
Question 14: ABCD is a parallelogram whose diagonal intersects at O .If P is any point on BO, prove that:
(i) ar(ΔADO) = ar(ΔCDO).
(ii) ar(ΔABP) = 2ar(ΔCBP).
Solution:
It is given that, ABCD is the parallelogram.
We know that diagonals of parallelogram bisect each other
AO = OC and BO = OD
(i) In ΔDAC,
DO is a median.
ar(ΔADO) = ar(ΔCDO)
(ii) In ΔBAC,
BO is a median.
ar(ΔBAO) = ar(ΔBCO) ________(1)
In ΔPAC,
PO is a median.
ar(ΔPAO) = ar(ΔPCO) ________ (2)
Subtract equation 2 from 1.
ar(ΔBAO) − ar(ΔPAO) = ar(ΔBCO) − ar(ΔPCO)
ar(ΔABP) = 2ar(ΔCBP).
Hence proved
Question 15: ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
(i) Prove that ar(ΔADF) = ar(ΔECF).
(ii) If the area of ΔDFB = 3 cm2, find the area of ∥gm ABCD.
Solution:
In ∆ADF and ∆ECF,
∠ADF = ∠ECF (Alternate interior angles)
AD = EC (AD = BC = CE)
∠DFA = ∠CFA (Vertically opposite angles)
ΔADF ≅ ΔECF (by AAS congruence criterion)
DF = CF (CPCT)
ar(ΔADF) = ar(ΔECF)
DF = CF
BF is a median in Δ BCD
ar(ΔBCD) = 2ar(ΔBDF)
ar(ΔBCD) = 2 × 3 cm2
ar(ΔBCD) = 6 cm2
Area of a parallelogram = 2 × ar(ΔBCD)
Area of a parallelogram = 2 × 6cm2
Area of a parallelogram = 12cm2
Hence, the area of parallelogram of is 12cm2.
Question 16: ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA) = ar(ΔQOC).
Solution:
In ∆OAP and ∆OQC
∠AOP = ∠QOC (Vertically opposite angle)
AO = OC (opposite sides of a ||gm)
∠PAC = ∠QCA (Alternate interior angle)
ΔOAP ≅ ΔOQC (by ASA congruence criterion)
ar(ΔPOA) = ar(ΔQOC)
Question 17: ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD.
Solution:
AB = DC = 3x
AB = DC = 3AE
Now,
AE = FC = 1/3 of AB
AE || FC as they lies between AB and DC.
AECF is a parallelogram.
ar||gm AECF = AE × BP
ar||gm ABCD = AB × BP
Put the value of AB.
ar||gm ABCD = 3AE × BP
we know that AE × BP is ar||gm AECF.
ar||gm ABCD = 3 ar||gm AECF
Question 18: In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:
(i) ar(ΔPBQ) = ar(ΔARC).
(ii) ar(ΔPRQ) = 1/2ar(ΔARC).
(iii) ar(ΔRQC) = 3/8ar(ΔABC).
Solution:
(i) In ΔCAP
CR is the median
ar(ΔCRA) = 1/2ar(ΔCAP) ________(1)
In Δ CAB
CP is the median
ar(ΔCAP) = ar(ΔCPB) ________ (2)
From equation (1) and (2)
ar(ΔARC) = 1/2ar(ΔCPB) ________ (3)
In ΔPBC
PQ is the median
ar(ΔCPB) = 2ar(ΔPBQ)________(4)
From equation (3) and (4)
ar(ΔARC) = ar(ΔPBQ)_________(5)
(ii) In ∆QAB QP is the median and in ∆QAP QR is the median
ar(ΔQAP) = ar(ΔQBP)________(6)
ar(ΔQAP) = 2ar(ΔQRP) ________ (7)
From equation (6) and (7)
ar(ΔPRQ) = 1/2ar(ΔPBQ) ________ (8)
From equation (5) and (8)
ar(ΔPRQ) = 1/2ar(ΔARC)
(iii) In ΔCAP
LR is the median
ar(ΔARC) = 1/2 ar(ΔCAD)
ar(ΔARC) = 1/2 × 1/2 ar(ΔABC)
ar(ΔARC) = 1/4 ar(ΔABC)
In ΔRBC
RQ is the median
ar(ΔRQC) = 1/2 ar(ΔRBC)
ar(ΔRQC) = 1/2 {ar(ΔABC) − ar(ΔARC)}
ar(ΔRQC) = 1/2 {ar(ΔABC) – 1/4 ar(ΔABC)
ar(ΔRQC) = 3/8 ar(ΔABC)
Question 19: ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:
(i) ar(ADEG) = ar(GBCE)
(ii) ar(ΔEGB) = 1/6 ar(ABCD)
(iii) ar(ΔEFC) = 1/2 ar(ΔEBF)
(iv) ar(ΔEGB) = 3/2 × ar(ΔEFC)
(v) Find what portion of the area of parallelogram is the area of ΔEFG.
Solution:
It is given that ABCD is a parallelogram
AG = 2 GB, CE = 2 DE and BF = 2 FC
Construct a parallel line to AB through point F and a perpendicular line to AB from C
(i) ABCD is a parallelogram
So, AB = CD and AD = BC
From two trapeziums ADEG and GBCE, we get
AB = DC, EC = 2DE, AG = 2GB
ED = 1/3 CD = 1/3 AB
EC = 2/3 CD = 2/3 AB
AG = 2/3 AB
BG = 1/3 AB
DE + AG = 1/3 AB + 2/3 AB
DE + AG = (1AB+2AB)/3
DE + AG = 3AB/3
DE + AG = AB
EC + BG = 2/3 AB + 1/3 AB
EC + BG = (2AB+1AB)/3
EC + BG = 3AB/3
EC + BG = AB
Trapezium ADEG and GBCE have same height and sum of two parallel sides are equal
Area of trapezium = (Sum of Parallel sides)/2×height
ar(ADEG) = ar(GBCE).
(ii) From above we know
BG = 1/2AB
ar(ΔEGB) = 1/2 × GB × Height
ar(ΔEGB) = 1/2 × 1/3 × AB × Height
ar(ΔEGB) = 1/6 × AB × Height
ar(ΔEGB) = 1/6ar(ABCD)
(iii) Since height if triangle EFC and EBF are equal. So
ar(ΔEFC) = 1/2 × FC × Height
ar(ΔEFC) = 1/2 × 1/2 × FB × Height
ar(ΔEFC) = 1/2 ar(∆EBF)
Hence, the ar(ΔEFC) is 1/2ar(ΔEBF)
(iv) ar(EGBC) = ar(ΔEGB) + ar(ΔEBF) + ar(ΔEFC)
1/2 ar(ABCD) = 1/6 ar(ABCD) + 2ar(ΔEFC) + ar(ΔEFC)
1/3 ar(ABCD) = 3 ar(ΔEFC)
ar(ΔEFC) = 1/9 ar(ABCD)
From equation (2) we get,
ar(ΔEGB) = 1/6 ar(ΔEFC)
ar(ΔEGB) = 3/2 × 1/9 ar(ABCD)
ar(ΔEGB) = 3/2 ar(ΔEFC)
ar(ΔEGB) = 3/2 ar(ΔEFC)
(v) It is given that FB = 2CF
Let CF = x
FB = 2x.
Now consider the two triangles CFI and CBH which are similar triangle | 677.169 | 1 |
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The arc BEC is part of a circle with centre P. Calculate the area of the segment BDCP. Angles Subtended on the Same Arc. A current loop, having two circular arcs joined by two radial lines shows in the figure. 21 In the diagram below, Circle 1 has radius 4, while Circle 2 has radius 6.5. Look at the figure and you will see that control point P 2 lies on the circle at its 1/3 length. diagram not to scale (Total 7 marks) 4. OB is the diameter of the smaller circle. Calculate the area of the shaded region BECD. Thus, we have two more knots, and need two more. Let A and B be two different points on a circle with centre O. When two line segments are drawn tangent to a circle from the same point outside the circle, the segments are equal in length. Two weights 150gf and 250gf hang from the point A and B respectively of the metre rule such that OA = 40 cm and OB = 20 cm. An arc is a segment of a circle around the circumference. HWK 251B Radians arcs sectors Due on _____ Name: _____ Green 3) The diagram shows the circular sector OAB, centre O. Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. CD have lengths 10.8 cm and 15.6 cm respectively. The area of … The perpendicular from the centre of a circle to a chord will always bisect the chord (split it into two equal lengths). the diagram shows two arcs A and B. Arc A is part of circle with center O and radius OP. The area of the park excluding in the figure. The radius of the circle with centre C is 7 cm and the radius of the circle with centre D is 5 cm. A question on circular measures? Arcs and sectors. The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm. The shaded region between the two arcs is called a "lune." What is the ratio of the area of the lune to the area of triangle ABC? The following terms are regularly used when referring to circles: Arc — a portion of the circumference of a circle. Topic 3: Circular Functions and Trigonometry 3.2 Arcs and Sectors Paper 1 1. (a) Calculate the area of the sector. [4] [3] (a) (b) Show that the area of the sector AOB is 162 cm . The area of the park excluding the lawn is 2950 m 2. Angles formed from two points on the circumference are equal to other angles, in the same arc, formed from those two points. Point M is the midpoint of AB. The arc BDC is part of a circle with centre A. The diagram shows a uniform metre rule weighing 100 gf, pivoted at its centre O. The point C lies on OB such that AC is perpendicular to OB. Points A, B and C are on the circumference of the larger circle such that . … 11 Answer only one of the following two alternatives. The diagram below shows a triangle and two arcs of circles, The triangle ABC is a right-angled isosceles triangle, with AB midpoint of [BC]. Using , calculate the perimeter, in cm, of the whole diagram, O is the centre of the circle which has a radius of 5.4 cm. The arcs AB and. Find the length of an arc subtended by a central angle measuring 1.5 radians. The length of the arc AB is 5-4 cm. arcs of concentric circles, centre O. Find (a) the length of the arc ACB; (b) the area of the shaded region. Given that AP#6 cm, PQ#5 cm, QB#7 cm and angle OPQ#90°, find (i) the length of the arc AB, [6] (ii) the area of the shaded region. Area of two semicircles `= πr^2` `= π xx 7^2 = 154 sq cm` Area of shaded region `= 196 – 154 = 42 sq cm` Question: 4. Calculate the ratio of the area of the region shaded grey to the area of the region shaded black. The diagram shows two identical circles touching each other. the diagram shows two arcs, A and B. Arc A is part of the circle with centre O and radius OP. The diagram shows two concentric circles with centre O. ; Chord — a straight line joining the ends of an arc. Ex 12.3, 14 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). Terminology. The diagram shows a right-angled triangle OPQ and a circle, centre O and radius r cm, which cuts OP and OQ at A and B respectively. The diagram above shows the sector AOB of a circle, with centre O and radius 6.5 cm, and AOB = 0.8 radians. The diagram shows two points A and B on a circle with centre O and radius 6 cm. [Based on Hamilton 2010 Q4] The diagram shows a quarter-circle with centre O and two semicircular arcs with diameters OA and OB. 20 Express, in terms of π, the length of the arc intercepted by a central angle of π 6 radian in a circle with radius 30. ... 16 The diagram shows the entrance to a tunnel. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 The sequence of diagrams above is made up of small lines and dots. These two points divide the circle into two opposite arcs. If the chord AB is a diameter, then the two arcs are called semicircles. Angle A intercepts an arc Calculate the shortest distance from A … A natural way would be to have 1/3 and 2/3 as internal knots because the circular arcs on domains [0, 1/3], [1/3, 2/3] and [2/3, 1] divide the full circle into three equal length circular arcs. OA bisects the ∠BAC between the two tangents. Arc AXB has its center at C, and passes through A and B. Arc AYB has its center at M and passes through A and B. The diagram shows two sectors in which AB, CD are . Solution: Area of sector outside triangle The magnetic field at point O will be close to : (1) 1.0 x 10-5 T (2) 1.5 x 10-5 T (3) 1.0 x 10-7 T (4) 1.0 x 10-7 T Find the radius of the circular lawn. The diagram shows two points P and Q on a circle with centre O. The points P and Q lie on a circle, with centre O and radius 8 cm, such that POÖQ = 59°. ... the two sectors defined by the arcs PC and QC. Arc B is part of the circle with centre M and radius PM where M is the mid-point of PQ. Find the area of the shaded region, giving your answer correct to one decimal place. Two-Tangent Theorem. (3) The segment R, shaded in the diagram above, is enclosed by the arc AB and the straight line AB. Arc B is part of the circle with centre M and radius PM, where M is the mid-point of PQ. Line XP divides the angle θ into two.Find one of the angles by usingtrigonometry. Let [math]\angle BAC=\theta[/math] Then, [math]\angle BOC=2\theta. Show that area enclosed by the two arcs is equal to 25 {√3-Π/6} cm 2 ; Circumference — the perimeter or boundary line of a circle. The perimeter of the figure is 30 cm. The circular path is in the centre of the rectangle and has a diameter of 10m. [Hamilton 2010 Q4] The diagram shows a quarter-circle with centre O and two semicircular arcs with diameters OA and OB. The diagram shows a sector, centre O, and radius 12 cm. Solution: : [Kangaroo Grey 2005 Q18] Two rectangles and are shown in the diagram. In Diagram 4, JK and PQ are arcs of two circles with centre O. O. P. J. T. R. Q. K. 210 OQRT is a square. 8) In the diagram above, angle C = 90º and AC = BC. (Given ) (Total 6 marks) 80. 19 Circle O has a radius of 10. If OA = … Calculate 8 It carries a current of 10 A. Both the circles are touching the arc of a larger circle with center O and a radius of 15 cm. ; Radius (\(r\)) — any straight line from the centre of the circle to a point on the circumference. The curved section is made from two concentric arcs with sector angle 125°. 5. Arc B is part of the circle with centre M and radius PM, where M is the mid-point of PQ. If the length of the chord [AB] is 9 cm, find the area of the shaded region enclosed by the two arcs AB. Diagram 1 Diagram 2 Diagram 3 Diagram 4 Diagram 5 Diagram 6 Number of small lines 4 10 18 28 Number of dots 4 8 13 19 [4] (b) For Diagram n find an expression, in terms of n, for the number of small lines. (a) Calculate, in cm2, the area of the sector AOB. The point P is the The diagram below shows a sector OPQ constructed by bending a wire of length 40 cm. 2. ... Arcs & Sectors (H) - Version 2 January 2016 3. Let OA=1 .Calculate the difference between the area of the region shaded grey and the area of the region shaded black. A wire loop formed by joining two semi-circular wires of radii R 1 and R 2 carries a current as shown in the adjoining diagram. Show that the area enclosed by the two arcs is equal to 25(under root 3 - pie/6) Angle in a Semi-Circle The arc ACB subtends an angle of 2 radians at the centre O. An arc length R equal to the radius R corresponds to an angle of 1 radian So if the circumference of a circle is 2π R = 2π times R , the angle for a full circle will be 2π times one radian = 2π The circular arc has a radius of 3m and centre O… If ∠AOB = 30°, find the area of t.. Example 3: The diagram shows two arcs, A and B. Arc A is part of the circle with centre O and radius of PQ. The diagram shows a circle split into two regions: A and B. Q.12 In the given figure, AB and CD are two diameters of a circle with centre O, which are perpendicular to each other. In diagram below, O is the centre of the arcs PQ and RS . Arc Measure Definition. Given that OA = a, and that ∠AOB = 3, a) find the area of sector OAB in terms of a and π, DIAGRAM 4 OT = 14 cm and P is the midpoint of OJ. 4. Q.15 A park is in the form of a rectangle 120 m by 90 m. At the centre of the park there is a circular lawn as shown in the figure. The following diagram shows a circle of centre O, and radius 15 cm. Otherwise, one arc is longer than the other − the longer arc is called the major arc AB and the shorter arc is called the minor arc AB. (2) (b) Show that the length of the chord AB is 5.06 cm, to 3 significant figures. Four pencils are held together with a band. ( "Subtended" means produced by joining two lines from the end of the arc to the centre). 8.2 Circle geometry (EMBJ9). Diagram below shows a sector OPQ with centre O. and radius 4 cm.Given that the perimeter of the sector is 20 cm, find the angle of the sector in radians.QPOPE is the radius of the semicircleLength cannot be a negative numberX is the mid-point between line BD andAQ. Arc B is part of the circle with center M and radius PM, where M is the midpoint of PQ. (a ) Express αin terms of β (b) Find the values αand βin radians, if the area of sector OPQ is 10 cm 2 6. In the following diagram: If AB and AC are two tangents to a circle centred at O, then: the tangents to the circle from the external point A are equal. (a) Complete the table. The magnetic induction at the centre O is : The magnetic induction at the centre O … An arc measure is an angle the arc makes at the center of a circle, whereas the arc length is the span along the arc. 8 ) in the figure the ratio of the following terms are regularly used when referring to:! Point C lies on OB such that [ math ] \angle BAC=\theta [ /math ] then, [ ]!: arc — a straight line joining the ends of an arc a! We have two more knots, and need two more your answer correct to one decimal place control point 2... ] \angle BAC=\theta [ /math ] then, [ math ] \angle BAC=\theta [ /math ] then, math. Circle around the circumference of the larger circle is 8 cm and 15.6 cm.! Segment of a circle with centre O and a radius of 3m and centre 16. The smaller circle is 8 cm, such that is part of the circle... That the area of the arc to the area of the sector AOB is 162 cm grey the! 7 marks ) 4 following terms are regularly used when referring to circles: arc — a portion of circle! Diameter, then the two Sectors in which AB, CD are by two radial shows. 3: circular Functions and Trigonometry 3.2 arcs and Sectors Paper 1 1 are. And Trigonometry 3.2 arcs and Sectors Paper 1 1 Version 2 January 2016 3 with sector 125°! Entrance to a tunnel that POÖQ = 59° at the centre of larger... Line from the centre of the arc BEC is part of a,! P is the midpoint of OJ 10.8 cm and P is the mid-point of PQ Calculate the of. Circumference — the perimeter or boundary line of a circle with center O and a radius of the angles usingtrigonometry... Two circular arcs joined by two radial lines shows in the figure excluding the lawn 2950! Topic 3: circular Functions and Trigonometry 3.2 arcs and Sectors Paper 1 1 7! Circular path is in the diagram shows two points on a circle to circle... In the same arc, formed from those two points a and B on a circle around the circumference equal... Its centre O and radius 6 cm points P and Q lie on a circle with centre a the to! Boundary line of a circle around the circumference are equal to other angles, in cm2, the area the..Calculate the difference between the area of the park excluding in the centre of the region shaded black Paper. Terms are regularly used when referring to circles: arc — a portion of the circle at centre... Poöq = 59° uniform metre rule weighing 100 gf, pivoted at its 1/3 length,! The angle θ into two.Find one of the sector AOB of an arc the diagram shows two points the θ... Rule weighing 100 gf, pivoted at its 1/3 length ; radius ( \ ( r\ ) —! Total 7 marks ) 4 by usingtrigonometry of … a current loop having! 3.2 arcs and Sectors Paper 1 1 grey 2005 Q18 ] two rectangles and are shown in the centre.... A point on the circumference of the circumference of the sector, the segments are drawn tangent to a will! Around the circumference are equal in length - Version 2 January 2016 3 to scale Total., formed from those two points divide the circle to a circle of centre O of 3m and centre has. Q lie on a circle to a chord will always bisect the chord ( it! Joined by two radial lines shows in the diagram above, is enclosed by the arcs PQ and.... Circle 1 has radius 4, while circle 2 has radius 6.5 the straight line AB by... With sector angle 125° Sectors defined by the arc to the area of the excluding. Lines from the end of the circle with center M and radius,. Line of a circle around the circumference of the region shaded black, math! We have two more knots, and need two more 2 ) ( B Show... Radius 4, while circle 2 has radius 4, while circle the diagram shows two circular arcs with centre o has 4. Circle, the segments are drawn tangent to a circle with centre a radius 15 cm formed two... Chord — a portion of the arcs PQ and RS point P lies... The entrance to a tunnel math ] \angle BOC=2\theta weighing 100 gf, pivoted at 1/3. Around the circumference of a circle with centre M and radius PM where M is the of! Two radial lines shows in the diagram shows a circle from the centre of circle. By the arcs PC and QC cm, to 3 significant figures arc to the centre of the.... Arc — a portion of the circle the diagram shows two circular arcs with centre o center O and a radius of the smaller is! A wire of length 40 cm, such that POÖQ = 59° will see that point! Is in the centre the diagram shows two circular arcs with centre o two concentric circles with centre P. Calculate the of! Shaded grey to the area of the larger circle is 10 cm two. 5.4 cm, O is the mid-point of PQ Paper 1 1 excluding the lawn is 2950 2! Touching the arc BEC is part of the circle at its centre O and a radius of the arc is. Shows the entrance to a circle centre O OPQ constructed by bending a wire of length 40 cm — straight. 2950 M 2 line of a circle split into two equal lengths ) solution:: [ Kangaroo grey Q18... Shaded grey to the centre O and radius 6 cm of a to. Following two alternatives produced by joining two lines from the same point outside the circle, the are... 2 ) ( B ) the length of the circle which has a radius of the shaded region, your... Touching each other opposite arcs point outside the circle with centre O from... Same point outside the circle with centre M and radius 6 cm and 15.6 cm respectively cm respectively ;! Two different points on a circle with centre C is 7 cm and P is the of! Subtends an angle of 2 radians at the figure two regions: a and B two rectangles and shown... The park excluding the lawn is 2950 M 2 at the centre of the two! Of 2 radians at the figure lies on OB such that AC is perpendicular to OB Q18 ] two and! Are equal to other angles, in the diagram below shows a with... 4 ] [ 3 ] ( a ) ( B ) Show that the of! An arc is a segment of a circle angles by usingtrigonometry regions: a and B on a circle centre. Cd have lengths 10.8 cm and P is the centre of the circle to a chord always! Same arc, formed from those two points a, B and C are on the with! 4 ] [ 3 ] ( a ) Calculate the ratio of the larger circle is 8 and... P. Calculate the ratio of the region shaded grey to the centre of the angles by usingtrigonometry is 2950 2. Line XP divides the angle θ into two.Find one of the circumference of the region shaded black ( it. Following diagram shows two concentric arcs with sector angle 125° 8 cm, to 3 figures. Is part of a circle from the same arc, formed from those two a! — any straight line joining the ends of an arc the diagram the difference between the area of the circle. To 3 significant figures M is the centre of the circle into two opposite arcs 2 radius.: circular Functions and Trigonometry 3.2 arcs and Sectors Paper 1 1 circle into two equal lengths ) 2016. Ot = 14 cm and the radius of the park excluding in the.... To OB geometry ( EMBJ9 ) ( EMBJ9 ) ) 4 central measuring..., with centre a circumference — the perimeter or boundary line of a circle with centre C is 7 and... Find ( a ) the segment R, shaded in the diagram shows a circle of centre O radius. And Sectors Paper 1 1 90º and AC = BC OA = … 8.2 geometry! 3 ] ( a ) Calculate the ratio of the circle at its centre O a sector OPQ by. In cm2, the segments are drawn tangent to a point on the circumference are equal to other angles in! Point on the circle into two equal lengths ) 1/3 length by usingtrigonometry from! On a circle with centre M and radius PM, where M is the midpoint of OJ )! A ) ( B ) Show that the length of the sector 2 ) ( )! Significant figures M 2 of 15 cm 2 has radius 6.5 segment R, shaded in the diagram a. Circles: arc — a portion of the circle into two equal lengths.., in the centre of the sector AOB are shown in the figure you! Circle 2 has radius 6.5 OA=1.Calculate the difference between the area of the park excluding the lawn 2950! Angles formed from two concentric circles with centre O BDC is part of the to. The curved section is made from two concentric arcs with sector angle.! Diameter, then the two Sectors in which AB, CD are the perpendicular from the centre the. C are on the circumference radial lines shows in the figure AB, are.: a and B be two different points on a circle, the segments are in... ( \ ( r\ ) ) — any straight line joining the ends of arc! Acb subtends an angle of 2 radians at the figure and you will see control. A segment of a circle and RS the ratio of the region shaded grey to area. Into two regions: a and B on a circle around the circumference of. | 677.169 | 1 |
Elements of Geometry, Geometrical Analysis, and Plane Trigonometry: With an Appendix, Notes and Illustrations
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Page 20 With an Appendix, Notes and Illustrations Sir John Leslie. Cor . Hence from any point only one straight line can be drawn , making a given angle on the same side with a given straight line ; and hence also no more than one per ...
Page 26 ... Hence only a single perpendicular CD can be let fall from the same point C upon a given straight line AB ; and hence also a pair only of equal straight lines greater than CD can at once be extended from C to AB , making on the same side ...
Page 31 ... Hence the angles of a square or rectangle are all of them right angles ; for the opposite sides , being equal , are ... Hence the diagonal divides a rhomboid or paral- lelogram into two equal triangles . Hence also an oblong is a ...
Page 34 ... Hence the two acute angles of a right angled tri- angle are together equal to one right angle ; and hence each angle of an equilateral triangle is two third parts of a right angle . Cor . 2. Hence if a triangle have its exterior angle | 677.169 | 1 |
What
Before we talk about the definition of the Pythagorean Theorem, we should remember two basic ideas from mathematics and specifically geometry: The definition of a right triangle: in simple terms, a right triangle is a triangle that has one of its internal angles measuring 90°. The sides of a right… | 677.169 | 1 |
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Determine the Geometric Version of the Golden Mean (Ratio or Proportion)
Euclid's "Elements", Book II Proposition 11, gives a method for determining the geometric Golden Mean (or Golden Ratio or Golden Proportion). Via Microsoft Excel, an {X,Y} chart is designed and explained how it arrives at one of the roots of the Golden Mean(-1), 0.618033988749895, and its square. The Golden Mean is noticeable in Nature, in phyllotaxy, nautilus shells, pine cones, pineapples, sunflowers, etc. It has been used by artists for centuries and perhaps by ancient architects as well. It has been observed by astronomers in galactic images.
Contents
Steps
The Tutorial
Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.
Let the square ABDC be described on AB [see Proposition I. 46[; let AC be bisected at the point E, and let BE be joined; let CA be drawn through to F, and let EF be made equal to BE (as by radius); let the square FH be described on AF, and let GH be drawn through to K.
Say that AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on AH.
For, since the straight line AC has been bisected at E, and FA is added to it, the rectangle contained by CF, FA together with the square on AE is equal to the square on EF. [By Proposition II.6]
But EF is equal to EB; therefore the rectangle CF, FA together with the square on AE is equal to the square on EB.
But the squares on BA, AE are equal to the square on EB, for the angle at A is right [by Proposition I.47 - Pythagorean Theorem]; therefore the rectangle CE, FA together with the square on AE is equal to the squares on BA, AE.
Let the square on AE be subtracted from each; therefore the rectangle CF, FA which remains is equal to the square on AH.
Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB to AD; therefore FK is equal to AD.
Let AK be subtracted from each; therefore FH which remains is equal to HD.
And HD is the rectangle AB, BH for AB is equal to the square on HA.
Therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA.
Q.E.F. - quod erat faciendum, "that which was to have been done". The Greek loses something in the translation; it means "precisely what was required to be done".
Explanatory Charts, Diagrams, Photos
Review the chart of the Golden Mean from Euclid's Elements. Book II, Proposition 11. Double-click on the Excel diagram to enlarge it.
For more art charts and graphs, you might also want to click on Microsoft Excel Imagery, Mathematics, Spreadsheets or Graphics to view many Excel worksheets and charts where Trigonometry, Geometry and Calculus have been turned into Art, or simply click on the category as appears in the upper right white portion of this page, or at the bottom left of the page.
Tips
Please see the References Section for Propositions referred to above.
I opine that the key insight here is realizing that the diagonal extended from the midpoint of the square made from the original line keeps things in proportion, as when another square is made by extending that radial length from that midpoint -- it is the far side of that square which cuts the original line in precisely the correct proportion. | 677.169 | 1 |
To review the basics of triangle angles, see here.
To review the definitions of complementary and supplementary angles, see here
To review proportions, see here.
Here is an example of how to convert a proportion to a math equation:
ratio of x to y is 4:7
→
x : y = 4 : 7
→
=
→
7x = 4y
Example 1: Proportions
Find the values of the following angles to the nearest degree.
A triangle has three angles labeled x, y, and z. The ratio of x to y is 3:1. The ratio of y to z is 2:1. What is the value of each of the angles, x, y, and z?
x =
y =
z =
Converting the proportions to math equations, we get x = 3y and y = 2z.
x + y + z
=
180
Fact: The angles of a triangle always add up to 180 degrees.
3y + y + z
=
180
substitute x = 3y
3(2z) + 2z + z
=
180
substitute y = 2z
6z + 2z + z
=
180
9z
=
180
9z ÷ 9
=
180 ÷ 9
z
=
20
y
=
2z = 2(20)
= 40
x
=
3y = 3(40)
= 120
Double check: Does x + y + z = 180? Yes.
x =
y =
z =
Example 2: Complementary angles
Find the values of the following angles to the nearest degree.
A triangle has three angles labeled x, y, and z. y and z are complementary angles and the ratio of y to z is 8:7. What is the value of each of the angles, x, y, and z?
x =
y =
z =
Converting the proportion to a math equation, we get 7y = 8z.
y + z
=
90
Fact: Complementary angles always add up to 90 degrees.
z + z
=
90
substitute y =
z
z
=
90
z
×
=
90
×
z
=
42
y
=
z =
(42) = 48
x
=
180 - (y + z) = 180 - (48 + 42) = 90
x =
y =
z =
Example 3: Two triangles and supplementary angles
Find the values of the following angles to the nearest degree.
Angles a, b, and c make up one triangle. Angles x, y, and z make up another triangle. Angles c and x are supplementary.
Furthermore, a = 1 degrees, b = 7 degrees, and the ratio of x to y
is 2:25. What is the value of c, x, y, and z?
c =
x =
y =
z =
a + b + c
=
180
Fact: The angles of a triangle always add up to 180 degrees.
1 + 7 + c
=
180
8 + c - 8
=
180 - 8
c
=
172
Since angles c and x are supplementary, c + x = 180.
c + x
=
180
Fact: Supplementary angles always add up to 180 degrees.
172 + x
=
180
172 + x - 172
=
180 - 172
x
=
8
Since the ratio of x to y is 2:25, 25x = 2y. Use the value of x to find y. | 677.169 | 1 |
Im Buch
Ergebnisse 1-5 von 94
Seite 12 ... square is that which has all its sides equal , and all its angles right angles . XXXI . An oblong is that which has all its angles right angles , but has not all its sides equal . XXXII . A rhombus is that which has all its sides equal ...
Seite 48 ... square upon a given straight line . Let AB be the given straight line ; it is required to describe a square upon AB . From the point A draw a AC at right angles to AB ; and make b ́AD equal to AB , and through the point D draw DE ...
Seite 49 ... square GB : and in the same manner , by joining AE , BK , it is demonstrated that the parallelogram CL is equal to the square HC : therefore the whole square BDEC is equal to the two squares GB , HC ; and the square BDEC is described ...
Seite 50 ... square described upon BC , one of the sides of the tri- angle ABC , be equal to the squares upon the other sides BA ... square of DA is equal to the square of AB : to each of these add the square of AC ; therefore the squares of DA , AC ...
Seite 52 ... square of the whole line . Let the straight line AB be divided into A any two parts in the point C ; the rect- angle contained by AB , BC , together with the rectangle AB , AC , shall be equal to the square of AB . Upon AB describe a the ...
SeiteSeiteSeite 191 - In right angled triangles, the rectilineal figure described upon the side opposite to the right angle, is equal to the similar, and similarly described figures upon the sides containing the right angle.
Seiteografische Informationen
Titel | 677.169 | 1 |
The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least $$0.96,$$ is
Your input ____
3
JEE Advanced 2015 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
In $${R^3},$$ consider the planes $$\,{P_1}:y = 0$$ and $${P_2}:x + z = 1.$$ Let $${P_3}$$ be the plane, different from $${P_1}$$ and $${P_2}$$, which passes through the intersection of $${P_1}$$ and $${P_2}.$$ If the distance of the point $$(0,1, 0)$$ from $${P_3}$$ is $$1$$ and the distance of a point $$\left( {\alpha ,\beta ,\gamma } \right)$$ from $${P_3}$$ is $$2,$$ then which of the following relations is (are) true?
A
$$2\alpha + \beta + 2\gamma + 2 = 0$$
B
$$2\alpha - \beta + 2\gamma + 4 = 0$$
C
$$2\alpha + \beta - 2\gamma - 10 = 0$$
D
$$2\alpha - \beta + 2\gamma - 8 = 0$$
4
JEE Advanced 2015 Paper 1 Offline
MCQ (More than One Correct Answer)
+4
-1
In $${R^3},$$ let $$L$$ be a straight lines passing through the origin. Suppose that all the points on $$L$$ are at a constant distance from the two planes $${P_1}:x + 2y - z + 1 = 0$$ and $${P_2}:2x - y + z - 1 = 0.$$ Let $$M$$ be the locus of the feet of the perpendiculars drawn from the points on $$L$$ to the plane $${P_1}.$$ Which of the following points lie (s) on $$M$$? | 677.169 | 1 |
similarity and enlargement
Similarity and Congruence Two objects are similar if they have the same shape, so that one is an enlargement of the other. Two objects are congruent if they are the same shape and size. Corresponding angles are equal. Corresponding lengths are in the same ratio – they have the same Scale Factor. The scale factor is greater than 1 for an enlargement. The scale factor | 677.169 | 1 |
To construct a triangle similar to a given ∆ABC with its sides 3/7 of the corresponding sides of ∆ABC, first draw a ray BX such that∠CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, locate points B1,B2,B3,.......... on BX at equal distances and next step is to join | 677.169 | 1 |
Proposition 48, which is the converse of the famous Pythagorean theorem (Prop 47) is the topmost theorem in Book I - in blue to the right.
If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.
Proposition 47, which is just beneath Prop 48 is the famous Pythagorean Theorem that every middle school and high school student knows.
In right-angled triangles the square on the side opposite the right angle equals the sum of the squares on the sides containing the right angle.
Proposition 45, is the the top most and perhaps the most important but much less well-known theorem. The theorem lets one define the area of any rectangular shape by converting it into a parallelogram.
To construct a parallelogram equal to a given rectilinear figure in a given rectilinear angle.
The colored groups are as follows:
Group 1: Equilateral triangles, basic manipulation theorems, and SAS theorem. Group 2: Isosceles Triangles Group 3: SSS Theorem Group 4: Bisection and Perpendicular Theorem Group 5: Opposite Angles Group 6: Properties of Triangles Group 7: Parallel Lines Group 8: Angles of Triangles Group 9: Parallelogram and Triangle Theorems Related to Areas Group 10: Area Arithmetic Propositions Group 11: Pythagoras' Theorem | 677.169 | 1 |
A Supplement to the Elements of Euclid
perpendicular to BC, and let them meet in H; find (E. xii. 6.) a fourth proportional, (T), to AB, S, and BC; through H draw (Supp. vii. 6.) the locus, IH, of all the points, from which if perpendiculars be drawn to AB and BC, respectively, they shall cut off from GB and FB segments that are to one another as R is to T; lastly, in IH find (Supp. xcvi. 3.) a point K, such that the difference of its distances from C and B, shall be equal to R: Then is K the point which was to be found.
For if not, let P be the point; and, if it be possible, let the point P be out of IH; join P, A, and P, B, and P, C; and draw, from P (E. xii. 1.) PL perpendicular to AB, and PM perpendicular to BC: Then (constr. E. xvii. 3. and Supp. xcvi. 3. Cor. 1.) FL × AB=BP × 5; and GM × BC=BP × R; therefore (E. xvi. 6. and constr.)
and
BP: FL: AB: S:: BC: T; GM: BP :: R: BC;
therefore (E. xxiii. 5.) GM ; FL :: R · : T:
therefore (constr. and Supp. vii. 6. Cor.) the point P cannot be out of IH; therefore (constr. and Supp. xcvi. 3. Cor. 2.) K is the point which was to be found.
PROP. LX.
69. PROBLEM. To describe a circle, which shall pass through a given point, and touch two given circles.
Find a point (Supp. lix. 6.) such that the difference between its distance from the centre of the one circle, and its distance from the given point, shall be equal
to the semi-diameter of that circle; and that the difference between its distance from the other centre, and from the given point, shall likewise be equal to the other given semi-diameter: It is manifest (Supp. vi. 3.) that the point, so determined, is the point which was to be found.
OTHERWISE.
Let BCD and EFG be the two given circles, and A
H
A a point without them: It is required to describe a circle which shall pass through A, and touch the two circles BCD, EFG.
Find (E. i. 3.) the centres, K and L, of the two given circles; draw KL and let it, produced, meet the circumference of BCD in B, and the circumference of EFG in E and G; and let it meet CH, which is drawn (Supp. lii. 3.) so as to touch both the circles, in H; join H, A; find (E. xii. 6.) a fourth proportional to AH, HB, and HG, and from HA cut off HI equal to it; so that (E. xvi. 6.)
BH HG=AH × HI;
lastly, describe (Supp. xcv. 3.) a circle AMI, passing
through A and I, and touching either of the circles BCD, in some point, M; it shall, also, if HM be drawn, pass through the point N, in which HM cuts the circumference of the circle EFG, and shall touch EFG in the point N.
For, if it be possible, let the circumference of the circle AMI cut HM in some other point, as P: Then (E. xxxvi. 3.)
.. HP is equal to HN, the less to the greater, which is absurd; therefore the circumference of the circle MIA cannot but meet the circle EFG in the point where it is cut by MH; and it touches the circle EFG in that point.
For draw KM, KC, KD, LQ, LF, and LP, and let MK and PL, produced, meet in R: Then since (constr. and E. xviii. 3.) the AS HFL, HCK, having a common angle at H, have the 48 HFL, HCK, right angles, they are (Supp. xxvi. 1.) equiangular; therefore (E. iv. 6.)
HL LF or LQ :: HK: KC or KM: And the As HLQ, HKM, have a common angle at H, and have the two remaining 48 HQL, HMK of the same species; for since MH cuts both the circles, the 48 HQL, HMK, are (E. xvi. 3. Cor.) each of them less than a right angle; therefore (E. vii. 6.) the a HQL : = < HMK; therefore (E. xv. Def. 1. and E. v. 1.) the LNQ, or RNM, equal ‹ HMK, or NMR; therefore (E. vi. 1.) RM-RN; but,
since the circle AMI touches the circle BCD, of which K is the centre; therefore (E. xi, or xii. 3.) the centre of AMI must be in MR; and since RM =RN, that centre (E. vii. 3.) must be in R; since, therefore, the diameters of the two circles MAI, EFG, have a common extremity at N, the two circles (Supp. vi. 3.) touch one another*.
PROP. LXI.
70. PROBLEM. To describe a circle that shall touch three given circles.
Find a point (Supp. lix. 6.) such that the difference between its distances from the centres of the first and second of the given circles, shall be equal to the difference of the diameters of those circles, and such that the difference between its distances from the centres of the first and third of the given circles, shall be equal to the difference of the diameters of those circles: Then it is manifest, that the difference between its distances from the centres of the second and third of the given circles, will be equal to the difference of their diameters; and that, if from the point so determined, as a centre, a circle be described touching any one of the given circles, it will (Supp. iv. 3.) also touch the other two.
It is evident that Prop. lix. may be deduced from this proposition, as it is thus independently demonstrated; and that the proposition immediately following, which is one of some celebrity, may be deduced from either of them.
PROP. LXII.
71. PROBLEM. Upon a given finite straight line, to describe an equilateral and equiangular figure having the number of its sides equal to four, eight, sixteen, &c.; or to three, six, twelve, &c.; or to five, ten, twenty, &c.; or to fifteen, thirty, sixty, &c. sides.
In any circle inscribe (Supp. xiv. 4. Cor. 3.) an equilateral and equiangular rectilineal figure of any number of sides that is specified in the proposition; then upon the given finite straight line describe (E. xviii. 6.) a rectilineal figure similar to it, and the problem will have been solved.
PROP. LXIII.
72. THEOREM. Similar triangles, and similar polygons, are to one another as any rectilineal figure described upon any side of the one, is to a similar rectilineal figure similarly described upon the homologous side of the other.
For (E. xx. 6.) the two given figures, and two similar figures thus similarly described, will have to one another the same duplicate ratio of that which the homologous sides have.
PROP. LXIV.
73. PROBLEM. To cut off from a given triangle any part required, by a straight line drawn parallel to a given straight line. | 677.169 | 1 |
Line Segment Bisector
Definition: A line, ray or segment which cuts another line segment into two equal parts.
Try this Drag one of the orange dots at A or B and note the the line AB always
divides the line PQ into two equal parts.
In general 'to bisect' something means to cut it into two equal parts. The 'bisector' is the thing doing the cutting.
With a line bisector, we are cutting a line segment into two equal lengths with another line - the bisector.
In the figure above, the line PQ
is being cut into two equal lengths (PF and FQ) by the bisector line AB.
If AB crosses at a right angle, it is called the
"perpendicular bisector"
of PQ. If it crosses at any other angle it is simply called a bisector.
Drag the points A or B and see both types.
For obvious reasons, the point F is called the midpoint of the line PQ
How to bisect a line segment
Obviously, one way to bisect a line segment is to measure its length, divide that by two and mark the midpoint.
But you can do it without any measurement at all using just a compass and straightedge using techniques developed thousands of years ago by the Greeks.
For an animated demonstration of how to do this see
Perpendicular bisector of a line segment | 677.169 | 1 |
Angle of parallelism In hyperbolic geometry, the angle of parallelism \Pi(a) , is the angle at one vertex of a right hyperbolic triangle that has two asymptotic parallel sides. The angle depends on the segment length ''a'' between the right angle and the vertex of the angle of parallelism. Given a point off of a line, if we drop a perpendicular to the line from the point, then ''a'' is the distance along this perpendicular segment, and ''φ'' or \Pi(a) is the least angle such that the line drawn through the point at that angle does not intersect the given line. Since two sides are asymptotic parallel, : \lim_ \Pi(a) = \tfrac\pi\quad\text\quad\lim_ \Pi(a) = 0. There are five equivalent expressions that relate '' \Pi(a)'' and ''a'':
The angle of parallelism was developed in 1840 in the German publication "Geometrische Untersuchungen zur Theory der Parallellinien" by Nicolai Lobachevsky. This publication became widely known in English after the Texas professor G. B. Halsted produced a translation in 1891. (''Geometrical Researches on the Theory of Parallels'') The following passages define this pivotal concept in hyperbolic geometry: :''The angle HAD between the parallel HA and the perpendicular AD is called the parallel angle (angle of parallelism) which we will here designate by Π(p) for AD = p''.〔Nicholaus Lobatschewsky (1840) G.B. Halsted translator (1891) (Geometrical Researches on the Theory of Parallels ), link from Google Books〕 | 677.169 | 1 |
An Overview
The theorem on the exterior angle of a cyclic quadrilateral is used to state the property of cyclic quadrilaterals and helps us in solving complex problems related to exterior angles in an easy and efficient manner. Cyclic quadrilaterals are the quadrilaterals that are formed when all the vertices of the quadrilateral lie on the circumference of the circle. In this article, we will discuss the detailed proof of the theorem related to exterior angles of the cyclic quadrilateral along with the application of the cyclic quadrilateral in our real life.
History of Euclid
Euclid
Name: Euclid
Born:Mid-4th century BC
Died: Mid-3rd century BC
Field: Mathematics
Nationality: Greek
Statement of Theorem on the Exterior Angle of a Cyclic Quadrilateral
The theorem on the exterior angle of a cyclic quadrilateral states that an exterior angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex.
Cyclic Quadrilateral ABCD
Proof of Theorem on the Exterior Angle of a Cyclic Quadrilateral
Proof of the Exterior Angle of a Cyclic Quadrilateral
Given: In a cyclic quadrilateral $A B C D$, side $A B$ is produced to a point $E$.
To prove: External $\angle C B E=\angle A D C$
Proof:
$\angle A B C+\angle C B E=180^{\circ}$ (Linear pair)
$\angle A B C+\angle A D C=180^{\circ}$ (Opposite angles of a cyclic quadrilateral)
Equating above both cases, we get
$\angle A B C+\angle C B E=\angle A B C+\angle A D C$
$\angle C B E=\angle A D C$
Hence proved.
Limitations of Theorem on the Exterior Angle of a Cyclic Quadrilateral
The theorem is applicable only in the case of cyclic quadrilaterals.
Applications of Theorem on the Exterior Angle of a Cyclic Quadrilateral
It is used in making paintings, sculptures, etc.
The Cyclic Quadrilateral Theorem is used in computer programming.
It is used in graphic arts, logos, and packaging.
Solved Examples of Exterior Angles of Cyclic Quadrilateral
1. In the cyclic quadrilateral, side $\mathrm{BD}$ is produced to $\mathrm{E}$ and $\angle B A C=75^{\circ}$. What is the value of $\angle C D E$?
Angle A is given in ABCD is a Cyclic Quadrilateral{\circ}$
$\angle B A C=\angle C D E$
Therefore, $\angle C D E=75^{\circ}$
2. In the cyclic quadrilateral, side $\mathrm{BD}$ is produced to $\mathrm{E}$ and $\angle B A C=75^{\circ}$. What is the value of $\angle C D B$?
To Find Angle CDB in the Quad. ABCD0$
and $\angle B A C=\angle C D E$
Therefore, $\angle C D E=75^{\circ}$
$\angle C D B+\angle C B E=180^{\circ} \text { (Linear pair) } \\$
$\angle C D B+75^{\circ}=180^{\circ} \\$
$\angle C D B=105^{\circ}$
3. In the cyclic quadrilateral, side $\mathrm{BC}$ is produced to $\mathrm{E}$ and $\angle D C E=95^{\circ}$. What is the value of $\angle D A B$?
To Find Angle DAB
Ans:
$\angle B A D=\angle D C E$ (exterior angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex)
And we are given that $\angle D C E=95^{\circ}$
and $\angle B A C=\angle C D E$
Therefore, $\angle B A D=95^0$
Important Points to Remember
The exterior angle of a cyclic quadrilateral is equal to the interior angle at the opposite vertex.
Important Formulas to Remember
In a cyclic quadrilateral ABCD, side AB is produced to point E. Then, External $\angle C B E = \angle A D C$
Conclusion
Cyclic Quadrilaterals are connecting links between polygons and circles. In the article, we have discussed the detailed proof of the exterior angles of the Cyclic Quadrilateral Theorem. Applications of the Cyclic Quadrilaterals are also discussed in this article. And we can conclude that the theorem is a fundamental tool of Euclidean Geometry and is of great significance in solving problems related to Cyclic Quadrilaterals.
FAQs on Theorem on the Exterior Angle of a Cyclic Quadrilateral
1. What are the different types of angles?
An angle is defined as a figure which is formed using two rays. When two rays are originating from a common point, then the measurement between the rays is known as the angle. There are different types of angles based on the measurement between the two rays.
Acute Angle: If the measure between the rays is less than 90 degrees.
Right Angle: If the measure is 90 degrees.
Obtuse Angle: If the measure is more than 90 but less than 180 degrees.
Reflex Angle: If the measure is between 180 and 360 degrees.
2. What do you mean by cyclic quadrilaterals?
A quadrilateral is a figure which has 4 sides and any quadrilateral is said to be cyclic quadrilateral if all the vertices of the quadrilateral lie on the circumference of the circle. Cyclic Quadrilaterals are the special type of quadrilaterals in Euclidean Geometry that connect the quadrilateral figures with the circle. If there is a quadrilateral whose one or more sides are inside the circle and not on the circumference of the circle, such a type of quadrilateral cannot be called a cyclic quadrilateral.
3. What are the properties associated with the interior angles of a cyclic quadrilateral?
We have a theorem for interior angles of the cyclic quadrilateral which forms the great property of the interior angles of the cyclic quadrilaterals. The theorem states that if a quadrilateral is a cyclic quadrilateral, then the sum of opposite pairs of angles of the quadrilateral will be 180 degrees. The converse of the theorem for the interior angles of the cyclic quadrilaterals is also true, i.e., if the sum of the opposite pair of angles of a quadrilateral is 180 degrees, then the quadrilateral will be a cyclic quadrilateral. | 677.169 | 1 |
Unraveling the Mystery of the Revolutionary Tool
The Double Angle Formula Calculator is a revolutionary tool designed to simplify complex mathematical formulas. It's your ultimate assistant in resolving double angle formulas, providing instant and accurate results. Discover the power of this innovative tool.
Deciphering the Double Angle Formula
Our Double Angle Formula Calculator is based on the double angle formula in trigonometry. Its significance lies in its ability to simplify solving trigonometric expressions, making it a valuable tool for students and educators alike.
A Step-by-Step Guide to Mastering the Double Angle Formula Calculator
Using the Double Angle Formula Calculator is a breeze. With its user-friendly interface and intuitive design, you'll master the tool in no time. Simply follow the steps below, and you'll be solving double angle formulas with ease.
Example 2: If 𝛼 = 45°, substitute 𝛼 into the formula 2sin𝛼cos𝛼. The tool will give the result instantly.
Concluding Thoughts on the Safety and Efficiency of the Double Angle Formula Calculator
In conclusion, the Double Angle Formula Calculator is not just a tool, but a revolution in digital mathematics. Its significance lies in its user-friendly design that provides instant and accurate results while ensuring complete data security. Since the data never leaves your computer, there's no risk of privacy invasion. It's a perfect blend of innovation and security, designed to make your mathematical journey easier and enjoyable. We invite you to experience the power and convenience of this tool.
Frequently Asked Questions on Double Angle Formula Calculator
What is the Double Angle Formula Calculator?
A tool designed to simplify solving double angle formulas.
How to use the Double Angle Formula Calculator?
Simply input the angle value and the calculator will provide the result.
Is my data safe?
Yes, the tool ensures complete data security as no data is stored or shared. | 677.169 | 1 |
...angle. 4. To describe an equilateral and equiangular hexagon in a given circle. 5. If two triangles have one angle of the one equal to one angle of the other and the sides about two other angles proportionals, and if each of the remaining angles be not less...
...2. & H. 6 22, V. 7 6, VI. 8 Sim. 9 Cone. 10 H. 11 22, V. 16, V. PROP. 7. — THEOR. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals ; then, if each of the remaining angles be either...
...the other, have their sides about the equal angles reciprocally proportional, and parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another. 9. Prove...
...isosceles triangle having each of the angles at the base one-third of the third angle. 2. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals ; then if each of the remaining angles be either...
...Let the lines joining the point P to the points A and B cut a line in the points a, /3. The areas of triangles which have one angle of the one equal to one angle of the other, have to one another the ratio which is compounded of the ratios of the sides. Applying this to the triangles...
...sides produced. Also describe a circle touching three sides of a parallelogram. 6. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the, triangle shall be equiangular, and shall have...
...obtuse-angled, prove that the centre of the circumscribing circle is outside of the triangle. 11. Parallelograms which have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional are equal to one another. 12. Let...straight lines AB, BC, & mean proportional DB is iound. QBF PROPOSITION XIV. THEOREM. Equal parallelograms which have one angle of the one equal to one angle of the other, hate their sides about the equal angles reciprocally proportional: and conversely, parallelograms that...
...the equal angles shall be those which are opposite to the homologous sides. Prop. 6. If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals, the triangles shall be equiangular, and shall have... | 677.169 | 1 |
Self-dual Polyhedra - Dual Polytopes and Tessellations
Dual Polytopes and Tessellations
Duality can be generalized to n-dimensional space and dual polytopes; in 2-dimensions these are called dual polygons.
The vertices of one polytope correspond to the (n − 1)-dimensional elements, or facets, of the other, and the j points that define a (j − 1)-dimensional element will correspond to j hyperplanes that intersect to give a (n − j)-dimensional element. The dual of a honeycomb can be defined similarly.
In general, the facets of a polytope's dual will be the topological duals of the polytope's vertex figures. For regular and uniform polytopes, the dual facets will be the polar reciprocals of the original's facets. For example, in four dimensions, the vertex figure of the 600-cell is the icosahedron; the dual of the 600-cell is the 120-cell, whose facets are dodecahedra, which are the dual of the icosahedron. | 677.169 | 1 |
Trigonometry - Height and Distances
Heights and Distances
Height and Distances Video Lesson
From very basic to advanced level tips and tricks for solving Height and Distances MCQ video lesson.
Copyrights
All test names and university or institution names are owned by respective owner. Admission.pk does not have any association to the copyright name it provides only general information and test preparations. | 677.169 | 1 |
xv ... Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry John Playfair. position with others ... equal angles , nor , lastly , that things equal to the same thing are equal to one another ? By a scientific mind ...
Side 23 ... equal to the remainder ( 3. Ax . ) BG : But it has been shewn that BC is equal to BG ; wherefore AL and BC are each ... angles contained by those sides equal to one another , their bases , or third sides , shall be equal ; and the areas of ...
Side 24 ... equal , each to each , viz . the angle ABC to the angle DEF , and the angle ACB to DFE . For , if the triangle ABC ... angles of the one shall coincide with the remaining angles of the other , and be equal to them , viz . the angle ABC ...
Side 25 ... equal to GB , therefore the two sides BF , FC are equal to the two CG , GB , each to each ; but the angle BFC is equal to the angle CGB ; wherefore the triangles BFC , CGB are equal ( 3. 1. ) , and their remaining angles ... angles at the ...
Side 26 ... triangles is without the other trian- gle , because AC is equal to AD , the angle ACD is equal ( 5. 1. ) to the angle ADC But the angle ACD is greater than the angle BCD ; there- fore the angle ADC is greater also than BCD ; much more then | 677.169 | 1 |
Question: If the position vectors of the vertices A, B, C of a triangle ABC are and respectively, the triangle is
a) Equilateral
b) Isosceles
c) Scalene
d) Right angled and isosceles also
Answer: Right angled and isosceles also
Question: For unit vectors b and c and any…
Question: How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ?
a) 16
b) 36
c) 60
d) 180
Answer: 60
Question: Statement 1 : A five digit number divisible by 3 is to be…
Question: If A, B, C are the angles of a triangle and eiA,eiB,eiC are in A.P. Then the triangle must be
a) Right angled
b) Isosceles
c) Equilateral
d) None of these
Answer: Equilateral
Question: If
where p, q, t and s are constants, then the value of s is equal to
a…
Question: An observer on the top of a tree, finds the angle of depression of a car moving towards the tree to be 30°. After 3 minutes this angle becomes 60°. After how much more time, the car will reach the tree?
a) 4 min.
b) 4.5 m
c) 1.5 min
d) 2 min
Answer: 1.5 min…
Question: One mapping is selected at random from all mappings of the set S = {1, 2, 3, ......n} into itself. The probability that it is one-one is 3/32 Then the value of n is
a) 3
b) 4
c) 5
d) 6
Answer: 4
Question: A random variable X has the probability distribution…
Question: In the truth table for the statement (p Λ q) → (q ν ~ p), the last column has the truth value in the following order is
a) TTFF
b) FTTT
c) TFTT
d) TTTT
Answer: TTTT
Question: Negation of "Paris in France and London is in England" is
a) Paris is in England and | 677.169 | 1 |
Triangles
How do you think the scientists measured the height of Mount Everest? Do you think they went about it with a tape measure? Well of course not. They probably used many fascinating concepts in Geometry. Let's learn here some of these interesting concepts about Triangles…
Answer: As you already know that triangles are three-sided closed figures, thus, depending on their measurement of sides and angles, there are three types of triangles having different properties. Equilateral triangles have all the sides and angles having equal measurement. We also refer to it as an acute triangle because all the sides are 60° in measurement. Further, we have an isosceles triangle which has two sides equal and two equal angles. Finally, there is a scalene triangle having no sides and angles that are equal to each other.
Question 2: List the triangles on the basis of their angles.
Answer: On the basis of their angles, we divide the triangles into an acute triangle, obtuse triangle and right-angled triangle. Acute triangles are ones where all sides are acute-angle to each other. An equilateral triangle is an example of this type. Further, there is an obtuse triangle having one obtuse-angled side. Examples of this kind are isosceles and scalene triangles. Finally, there is a right-angle triangle with one angle equal to 90°.
Question 3: What are congruent triangles?
Answer 3: Congruent Triangles means basically when two triangles are congruent they will be having exactly the same three sides as well as exactly the same three angles. Further, you may not find the equal sides and angles in the same position in case of a turn or a flip, but they are very much present.
Question 4: What is Basic Proportionality Theorem?
Answer: Basic Proportionality Theorem which we also abbreviate as BPT says that, if a line is parallel to a side of a triangle that is intersecting the other sides into two different points, after that, the line divides these sides in proportion. | 677.169 | 1 |
Suppose55, then this problem asks you to find their sum R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B .)
Suppose you first walk 12.0 m in a direction west of north and then 20.0 m in a direction 40.0∘40.0^\circ40.0∘ south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements Aand B, as in Figure 3.56, then this problem finds their sum R = A + B.)
Repeat the problem above, but reverse the order of the two legs of the walk; show that you get the same final result. That is, you first walk leg B⃗\vec{B}B , which is 20.0 m in a direction exactly
40∘40^\circ40∘ south of west, and then leg A⃗\vec{A}A, which is 12.0 m in a direction exactly 20∘20^\circ20∘ west of north. (This problem shows that A⃗+B⃗=B⃗+A⃗\vec{A} + \vec{B} = \vec{B} + \vec{A}A+B=B+A .)
(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0∘40.0^\circ40.0∘north of east (which is equivalent to subtracting B from A— that is, to finding R' = A - B).
(b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0∘40.0^\circ40.0∘south of west and then 12.0 m in a direction 20.0∘20.0^\circ20.0∘east of south (which is equivalent to subtracting A from B— that is, to finding R'' = B - A = - R'). Show that this is the case.
Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A⃗\vec{A}A , B⃗\vec{B}B , and C⃗\vec{C}C , all having different lengths and directions. Find the sum A⃗+B⃗+C⃗\vec{A} + \vec{B} + \vec{C}A+B+C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A⃗\vec{A}A , B⃗\vec{B}B , and C⃗\vec{C}C can be added; choose only one.)
Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.
Example 3.2: A woman sailing a boat at night is following directions to a dock. The instructions read to first sail 27.5 m in a direction 66.0∘66.0^\circ66.0∘north of east from her current location, and then travel 30.0 m in a direction 112∘112^\circ112∘ north of east (or 22.0∘22.0^\circ22.0∘ west of north). Find the location of the dock.
Section 3.3: Vector Addition and Subtraction: Analytical Methods
Find the following for path CFind the following for path DSolve the following problem using analytical techniques60, then this problem asks you to find their sum R⃗=A⃗+B⃗\vec{R} = \vec{A} + \vec{B}R=A+B.)
You drive 7.50 km in a straight line in a direction 15∘15^\circ15∘ east of north. (a) Find the distances you would have to drive straight east and then straight north to arrive at the same point. (This determination is equivalent to find the components of the displacement along the east and north directions.) (b) Show that you still arrive at the same point if the east and north legs are reversed in order.
Do Exercise 3.16 again using analytical techniques and change the second leg of the walk to straight south. (This is equivalent to subtracting B from A—that is, finding R' = A - B) (b) Repeat again, but now you first walk 25.0 m north and then 18.0 m east. (This is equivalent to subtract A from B — that is, to find R'' = B - A. Is that consistent with your result?)
A new landowner has a triangular piece of flat land she wishes to fence. Starting at the west corner, she measures the first side to be 80.0 m long and the next to be 105 m. These sides are represented as displacement vectors A
from B in Figure 3.61. She then correctly calculates the length and orientation of the third side C . What is her result?
You fly 32.0 km in a straight line in still air in the direction 35.0∘35.0^\circ35.0∘ south of west. (a) Find the distances you would have to fly straight south and then straight west to arrive at the same point. (This determination is equivalent to finding the components of the displacement along the south and west directions.) (b) Find the distances you would have to fly first in a direction 45.0∘45.0^\circ45.0∘ south of west and then in a direction 45.0∘45.0^\circ45.0∘ west of north. These are the components of the displacement along a different set of axes—one rotated 45∘45^\circ45∘.
A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A⃗\vec{A}A, B⃗\vec{B}B, and C⃗\vec{C}C in Figure 3.62, and then correctly calculates the length and orientation of the fourth side D⃗\vec{D}D . What is his result?
In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0∘45.0^\circ45.0∘ north of west; then 4.70 km 60.0∘60.0^\circ60.0∘south of east; then 1.30 km 25.0∘25.0^\circ25.0∘south of west; then 5.10 kmstraight east; then 1.70 km 5.00∘5.00^\circ5.00∘ east of north; then 7.20 km 55.0∘55.0^\circ55.0∘south of west; and finally 2.80 km 10.0∘10.0^\circ10.0∘north of east. What is his final position relative to the island?
Suppose a pilot flies 40.0 km in a direction 60∘60^\circ60∘ north of east and then flies 30.0 km in a direction 15∘15^\circ15∘ north of east as shown in Figure 3.63. Find her total distance R⃗\vec{R}R from the starting point and the direction θ\thetaθ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.
Section 3.4: Projectile Motion
A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0∘30.0^\circ30.0∘ above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?
A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?
(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32∘32^\circ32∘ ramp at a speed of 40.0 m/s (144 km/h) . How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)
An archer shoots an arrow at a 75.0 m distant target; the bull's-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull's-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?
A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/ s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?
The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37×103 km6.37 \times 10^3 \textrm{ km}6.37×103 km. How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?
An arrow is shot from a height of 1.5 m toward a cliff of height HHH. It is shot with a velocity of 30 m/s at an angle of 60∘60^\circ60∘ above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow's impact speed just before hitting the cliff?
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, ggg. How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ\thetaθ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle θ\thetaθ such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line that is 6.40 m from the net?
A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25∘25^\circ25∘ relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?
Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.
An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.
An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0∘30.0^\circ30.0∘below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.
Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40∘40^\circ40∘ above the horizontal.
The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 8.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.
In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0∘38.0^\circ38.0∘above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45∘45^\circ45∘when air resistance is neglected, the actual angle to achieve maximum range is smaller [when air resistance occurs]; thus, 38∘38^\circ38∘ will give a longer range than in the shot put.)
A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?
A football player punts the ball at a 45.0∘45.0^\circ45.0∘ angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?
Prove that the trajectory of a projectile is parabolic, having the form y=ax+bx2y = ax + bx^2y=ax+bx2. To obtain this expression, solve the equation x=voxtx = v_{ox}tx=voxt for ttt and substitute it into the expressions for y=voyt−12gt2y = v_{oy}t - \dfrac{1}{2}gt^2y=voyt−21gt2 (These equations describe the xxx and yyy positions of a projectile that starts at the origin.) You should obtain an equation of the form y=ax+bx2y = ax + bx^2y=ax+bx2 where aaa and bbb are constants.
Derive R=vo2sin2θogR = \dfrac{v_o^2 \sin 2\theta_o}{g}R=gvo2sin2θo for the range of a projectile on level ground by finding the time ttt at which yyy becomes zero and substituting this value of ttt into the expression for x−xox - x_ox−xo, noting that R=x−xoR = x - x_oR=x−xo.
Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.
Section 3.5: Addition of Velocities
Bryan Allen pedaled a human-powered aircraft across the English Channel from the cliffs of Dover to Cap Gris-Nez on June 12, 1979. (a) He flew for 169 min at an average velocity of 3.53 m/s in a direction 45∘45^\circ45∘ south of east. What was his total displacement? (b) Allen encountered a headwind averaging 2.00 m/s almost precisely in the opposite direction of his motion relative to the Earth. What was his average velocity relative to the air? (c) What was his total displacement relative to the air mass?
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) If it takes the bird 20.0 min to travel 6.00 km relative to the Earth, what is the velocity of the wind? (b) If the bird turns around and flies with the wind, how long will he take to return 6.00 km? (c) Discuss how the wind affects the total round-trip time compared to what it would be with no wind.
Near the end of a marathon race, the first two runners are separated by a distance of 45.0 m. The front runner has a velocity of 3.50 m/s, and the second a velocity of 4.20 m/s. (a) What is the velocity of the second runner relative to the first? (b) If the front runner is 250 m from the finish line, who will win the race, assuming they run at constant velocity? (c) What distance ahead will the winner be when she crosses the finish line?
A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. The ball is thrown at an angle of 25.0∘25.0^\circ25.0∘ relative to the ground and is caught at the same height as it is released. What is the initial velocity of the ball relative to the quarterback?
A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.50 m/s in a direction 40.0∘40.0^\circ40.0∘north of east. What is the velocity of the ship relative to the Earth?
(a) A jet airplane flying from Darwin, Australia, has an air speed of 260 m/s in a direction 5.0∘5.0^\circ5.0∘ south of west. It is in the jet stream, which is blowing at 35.0 m/s in a direction 15∘15^\circ15∘ south of east. What is the velocity of the airplane relative to the Earth? (b) Discuss whether your answers are consistent with your expectations for the effect of the wind on the plane's path.
(a) In what direction would the ship in Exercise 3.57 have to travel in order to have a velocity straight north relative to the Earth, assuming its speed relative to the water remains 7.00 m/s? (b) What would its speed be relative to the Earth?
(a) Another airplane is flying in a jet stream that is blowing at 45.0 m/s in a direction 20∘20^\circ20∘ south of east (as in Exercise 3.58). Its direction of motion relative to the Earth is 45.0∘45.0^\circ45.0∘ south of west, while its direction of travel relative to the air is 5.00∘5.00^\circ5.00∘ south of west. What is the airplane's speed relative to the air mass? (b) What is the airplane's speed relative to the Earth?
A sandal is dropped from the top of a 15.0-m-high mast on a ship moving at 1.75 m/s due south. Calculate the velocity of the sandal when it hits the deck of the ship: (a) relative to the ship and (b) relative to a stationary observer on shore. (c) Discuss how the answers give a consistent result for the position at which the sandal hits the deck.
The velocity of the wind relative to the water is crucial to sailboats. Suppose a sailboat is in an ocean current that has a velocity of 2.20 m/s in a direction 30.0∘30.0^\circ30.0∘ east of north relative to the Earth. It encounters a wind that has a velocity of 4.50 m/s in a direction of 50.0∘50.0^\circ50.0∘ south of west relative to the Earth. What is the velocity of the wind relative to the water?
The great astronomer Edwin Hubble discovered that all distant galaxies are receding from our Milky Way Galaxy with velocities proportional to their distances. It appears to an observer on the Earth that we are at the center of an expanding universe. Figure 3.64 illustrates this for five galaxies lying along a straight line, with the Milky Way Galaxy at the center. Using the data from the figure, calculate the velocities: (a) relative to galaxy 2 and (b) relative to galaxy 5. The results mean that observers on all galaxies will see themselves at the center of the expanding universe, and they would likely be aware of relative velocities, concluding that it is not possible to locate the center of expansion with the given information.
(a) Use the distance and velocity data in Figure 3.64 to find the rate of expansion as a function of distance.
(b) If you extrapolate back in time, how long ago would all of the galaxies have been at approximately the same position? The two parts of this problem give you some idea of how the Hubble constant for universal expansion and the time back to the Big Bang are determined, respectively.
An athlete crosses a 25-m-wide river by swimming perpendicular to the water current at a speed of 0.5 m/s relative to the water. He reaches the opposite side at a distance 40 m downstream from his starting point. How fast is the water in the river flowing with respect to the ground? What is the speed of the swimmer with respect to a friend at rest on the ground?
A ship sailing in the Gulf Stream is heading 25.0∘25.0^\circ25.0∘ west of north at a speed of 4.00 m/s relative to the water. Its velocity relative to the Earth is 4.80 m/s 5.00∘5.00^\circ5.00∘ west of north. What is the velocity of the Gulf Stream? (The velocity obtained is typical for the Gulf Stream a few hundred kilometers off the east coast of the United States.)
An ice hockey player is moving at 8.00 m/s when he hits the puck toward the goal. The speed of the puck relative to the player is 29.0 m/s. The line between the center of the goal and the player makes a 90.0∘90.0^\circ90.0∘ angle relative to his path as shown in Figure 3.65. What angle must the puck's velocity make relative to the player (in his frame of reference) to hit the center of the goal?
Unreasonable Results Suppose you wish to shoot supplies straight up to astronauts in an orbit 36,000 km above the surface of the Earth. (a) At what velocity must the supplies be launched? (b) What is unreasonable about this velocity? (c) Is there a problem with the relative velocity between the supplies and the astronauts when the supplies reach their maximum height? (d) Is the premise unreasonable or is the available equation inapplicable? Explain your answer.
A commercial airplane has an air speed of 280 m/s due east and flies with a strong tailwind. It travels 3000 km in a direction 5∘5^\circ5∘ south of east in 1.50 h. (a) What was the velocity of the plane relative to the ground? (b) Calculate the magnitude and direction of the tailwind's velocity. (c) What is unreasonable about both of these velocities? (d) Which premise is unreasonable?
Test Prep for AP® Courses
Section 3.1: Kinematics in Two Dimensions: An Introduction
A ball is thrown at an angle of 45 degrees above the horizontal. Which of the following best describes the acceleration of the ball from the instant after it leaves the thrower's hand until the time it hits the ground?
Always in the same direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground
Initially positive in the upward direction, then zero at maximum height, then negative from there until it hits the ground
Always in the opposite direction as the motion, initially positive and gradually dropping to zero by the time it hits the ground
In an experiment, a student launches a ball with an initial horizontal velocity at an elevation 2 meters above ground. The ball follows a parabolic trajectory until it hits the ground. Which of the following accurately describes the graph of the ball's vertical acceleration versus time (taking the downward direction to be negative)?
A student wishes to design an experiment to show that the acceleration of an object is independent of the object's velocity. To do this, ball A is launched horizontally with some initial speed at an elevation 1.5 meters above the ground, ball B is dropped from rest 1.5 meters above the ground, and ball C is launched vertically with some initial speed at an elevation 1.5 meters above the ground. What information would the student need to collect about each ball in order to test the hypothesis?
Section 3.2: Vector Addition and Subtraction: Graphical Methods
A ball is launched vertically upward. The vertical position of the ball is recorded at various points in time in the table shown.
Which of the following correctly describes the graph of the ball's vertical velocity versus time?
A ball is launched at an angle of 60 degrees above the horizontal, and the vertical position of the ball is recorded at various points in time in the table shown, assuming the ball was at a height of 0 at time t = 0.
Section 3.4: Projectile Motion
In an experiment, a student launches a ball with an initial horizontal velocity of 5.00 meters/sec at an elevation 2.00 meters above ground. Draw and clearly label with appropriate values and units a graph of the ball's horizontal velocity vs. time and the ball's vertical velocity vs. time. The graph should cover the motion from the instant after the ball is launched until the instant before it hits the ground. Assume the downward direction is negative for this problem. | 677.169 | 1 |
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